$(1) binh phuong 2 ve\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loai) \vee y-8x+4=0\Rightarrow y-6x=2x-4 thay vao (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoa DK) \vee 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}=0 vo ly vi pt >0 \forall x \geq \frac{\sqrt{39}}{2}$

bình phương 2 vế pt(1): $4x-2\sqrt{4x^{2}-y}=8x-y\Leftrightarrow 2\sqrt{4x^{2}-y}=y-4x\Rightarrow 4(4x^{2}-y)=16x^{2}-4xy+y^{2}\Leftrightarrow y^{2}-8xy+4y=0\Rightarrow y=0(loai) \vee y-8x+4=0\Leftrightarrow y-6x=2x-4 thay vao pt (2) \sqrt{4x^{2}-39}=3+\sqrt{2x-4}\Rightarrow 4x^{2}-39=9+6\sqrt{2x-4}+2x-4\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2x^{2}-16x+32+15x-60+6-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}=0\Rightarrow x=4 \vee 2x+7-\frac{6}{\sqrt{2x-4}+2}\Leftrightarrow 2x+4+\frac{3\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{3\sqrt{2x-4}}{\sqrt{2x-4}+2}=0 vo ly do x\geq 2. vay voi x=4, y=28 thi thoa DK$

$ (1) binh phuong 2 ve\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loai) \vee y-8x+4=0\Rightarrow y-6x=2x-4 thay vao (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoa DK) \vee 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}=0 vo ly vi pt >0 \forall x \geq \frac{\sqrt{39}}{2}$

$(1) binh phuong 2 ve\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loai) \vee y-8x+4=0\Rightarrow y-6x=2x-4 thay vao (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoa DK) \vee 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}=0 vo ly vi pt >0 \forall x \geq \frac{\sqrt{39}}{2}$

$(1) binh phuong 2 ve\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loai) \vee y-8x+4=0\Rightarrow y-6x=2x-4 thay vao (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoa DK) \vee 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}=0 vo ly vi pt >0 \forall x \geq \frac{\sqrt{39}}{2}$

$ (1) binh phuong 2 ve\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loai) \vee y-8x+4=0\Rightarrow y-6x=2x-4 thay vao (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoa DK) \vee 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}=0 vo ly vi pt >0 \forall x \geq \frac{\sqrt{39}}{2}$

$\sqrt{2x+\sqrt{y}}-\sqrt{2x-\sqrt{y}}=\sqrt{8x-y}\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loại) hoặc y-8x+4=0\Rightarrow y-6x=2x-4 thay vào (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoã đk) hoặc 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}>0$

$(1) binh phuong 2 ve\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loai) \vee y-8x+4=0\Rightarrow y-6x=2x-4 thay vao (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoa DK) \vee 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}=0 vo ly vi pt >0 \forall x \geq \frac{\sqrt{39}}{2}$

$\sqrt{2x+\sqrt{y}}-\sqrt{2x-\sqrt{y}}=\sqrt{8x-y}\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loại) hoặc y-8x+4=0\Rightarrow y-6x=2x-4 thay vào (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoã đk) hoặc 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}>0\forall x\geq \frac{\sqrt{39}}{2}$

$\sqrt{2x+\sqrt{y}}-\sqrt{2x-\sqrt{y}}=\sqrt{8x-y}\Leftrightarrow -2\sqrt{4x^{2}-y}=4x-y\Rightarrow 16x^{2}-4y=16x^{2}-8xy+y^{2}\Leftrightarrow y^{2}-8xy-4y=0\Rightarrow y=0(loại) hoặc y-8x+4=0\Rightarrow y-6x=2x-4 thay vào (2)\Rightarrow \sqrt{4x^{2}-39}=3+\sqrt{2x-4} (x\geq \frac{\sqrt{39}}{2})\Leftrightarrow 2x^{2}-x-22-3\sqrt{2x-4}=0\Leftrightarrow 2(x-4)^{2}+15(x-4)-6.\frac{x-4}{\sqrt{2x-4}+2}\Rightarrow x=4, y=28(thoã đk) hoặc 2x+7-\frac{6}{\sqrt{2x-4}+2}=0\Leftrightarrow 2x+4+\frac{\sqrt{2x-4}+6-6}{\sqrt{2x-4}+2}>0$

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $(\sqrt{bc} \leq$ b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) $\leq$ a3 + ( 1-a)3 + \frac{left(4a-3(1-a)).(a-1)2right)}{4})$đặt f(a) = $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a $\in$ [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $(\sqrt{bc} \leq$ b+c = 1 -a => bc ≥ (a-1)2 /4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) ≤ a3 + ( 1-a)3 + (4a-3(1-a)).(a-1)2 /4đặt f(a) = a3 + ( 1-a)3 + (4a-3(1-a)).(a-1)2 /4 với a thuộc [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $(\sqrt{bc} \leq)$b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a))$(\leq a3 + ( 1-a)3 + \frac{(4a-3(1-a)).(a-1)2}{4})$đặt f(a) =$a^{3}$+$( 1-a)^{3}$+$\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $(\sqrt{bc} \leq$ b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) $\leq$ a3 + ( 1-a)3 + \frac{left(4a-3(1-a)).(a-1)2right)}{4})$đặt f(a) =$a^{3}$+$( 1-a)^{3}$+$\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a $\in$ [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc} \leq$ b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) $\leq$ a3 + ( 1-a)3 + $\frac{(4a-3(1-a)).(a-1)2}{4}$đặt f(a) =$a^{3}$+$( 1-a)^{3}$+$\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $(\sqrt{bc} \leq)$b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a))$(\leq a3 + ( 1-a)3 + \frac{(4a-3(1-a)).(a-1)2}{4})$đặt f(a) =$a^{3}$+$( 1-a)^{3}$+$\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc}$ $\leq$b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a))$\leq$a3 + ( 1-a)3 +$\frac{(4a-3(1-a)).(a-1)2}{4}$đặt f(a) =$a^{3}$+$( 1-a)^{3}$+$\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc} \leq$ b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) $\leq$ a3 + ( 1-a)3 + $\frac{(4a-3(1-a)).(a-1)2}{4}$đặt f(a) = $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a $\geq$ b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc}$ $\leq$ b+c = 1 -a => bc $\leq$ ($(a-1)^{2}$)/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) $\leq$ a3 + ( 1-a)3 + $\frac{(4a-3(1-a)).(a-1)2}{4}$đặt f(a) = $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a = b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc}$ $\leq$ b+c = 1 -a => bc ≥ (a-1)2/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) $\leq$ a3 + ( 1-a)3 + $\frac{(4a-3(1-a)).(a-1)2}{4}$đặt f(a) = $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a $\geq$ b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc}$ $\leq$ b+c = 1 -a => bc $\leq$ ($(a-1)^{2}$)/4$a^{3}$ + $b^{3}$ + $c^{3}$ + 4abc = $a^{3}$ + ( b+c)($b^{2}$ -bc + $c^{2}$) + 4abc= $a^{3}$ + ( b+c)($(b+c)^{2}$ - 3bc) + 4abc = $a^{3}$ + $( b+c)^{3}$ -3bc(b+c) + 4abc = $a^{3}$ + $( b+c)^{3}$ + bc(4a-3(1-a)) $\leq$ $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$đặt f(a) =$a^{3}$+$( 1-a)^{3}$+$\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a $\geq$ b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc}$ $\leq$ b+c = 1 -a => bc $\leq$ ($(a-1)^{2}$)/4a3 + b3 + c3 + 4abc = a3 + ( b+c)( b2 -bc + c2 ) + 4abc= a3 + ( b+c)((b+c)2 - 3bc) + 4abc = a3 + ( b+c)3 -3bc(b+c) + 4abc = a3 + ( b+c)3 + bc(4a-3(1-a)) $\leq$ a3 + ( 1-a)3 + $\frac{(4a-3(1-a)).(a-1)2}{4}$đặt f(a) = $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a $\geq$ b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2.$sqrt{bc}$ $\leq$ b+c = 1 -a => bc \leq ((a-1)^{2})/4a^{3} + b^{3} + c^{3} + 4abc = a^{3} + ( b+c)(b^{2} -bc + c^{2}) + 4abc= a^{3} + ( b+c)((b+c)^{2} - 3bc) + 4abc = a^{3} + ( b+c)^{3} -3bc(b+c) + 4abc = a^{3} + ( b+c)^{3} + bc(4a-3(1-a)) \leq a^{3} + ( 1-a)^{3} + \frac{(4a-3(1-a)).(a-1)^{2}}{4}đặt f(a) = a^{3} + ( 1-a)^{3} + \frac{(4a-3(1-a)).(a-1)^{2}}{4} với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a $\geq$ b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2. $\sqrt{bc}$ $\leq$ b+c = 1 -a => bc $\leq$ ($(a-1)^{2}$)/4$a^{3}$ + $b^{3}$ + $c^{3}$ + 4abc = $a^{3}$ + ( b+c)($b^{2}$ -bc + $c^{2}$) + 4abc= $a^{3}$ + ( b+c)($(b+c)^{2}$ - 3bc) + 4abc = $a^{3}$ + $( b+c)^{3}$ -3bc(b+c) + 4abc = $a^{3}$ + $( b+c)^{3}$ + bc(4a-3(1-a)) $\leq$ $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$đặt f(a) = $a^{3}$ + $( 1-a)^{3}$ + $\frac{(4a-3(1-a)).(a-1)^{2}}{4}$ với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a \geq b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2.sqrt{x}bc \leq b+c = 1 -a => bc \leq ((a-1)^{2})/4a^{3} + b^{3} + c^{3} + 4abc = a^{3} + ( b+c)(b^{2} -bc + c^{2}) + 4abc= a^{3} + ( b+c)((b+c)^{2} - 3bc) + 4abc = a^{3} + ( b+c)^{3} -3bc(b+c) + 4abc = a^{3} + ( b+c)^{3} + bc(4a-3(1-a)) \leq a^{3} + ( 1-a)^{3} + \frac{(4a-3(1-a)).(a-1)^{2}}{4}đặt f(a) = a^{3} + ( 1-a)^{3} + \frac{(4a-3(1-a)).(a-1)^{2}}{4} với a \in [0;1/2] rồi đạo hàm tìm max f(a)

ta có 1-a $\geq$ b+c, vì b,c > 0 nên ta áp dụng bdt côsi: 2.$sqrt{bc}$ $\leq$ b+c = 1 -a => bc \leq ((a-1)^{2})/4a^{3} + b^{3} + c^{3} + 4abc = a^{3} + ( b+c)(b^{2} -bc + c^{2}) + 4abc= a^{3} + ( b+c)((b+c)^{2} - 3bc) + 4abc = a^{3} + ( b+c)^{3} -3bc(b+c) + 4abc = a^{3} + ( b+c)^{3} + bc(4a-3(1-a)) \leq a^{3} + ( 1-a)^{3} + \frac{(4a-3(1-a)).(a-1)^{2}}{4}đặt f(a) = a^{3} + ( 1-a)^{3} + \frac{(4a-3(1-a)).(a-1)^{2}}{4} với a \in [0;1/2] rồi đạo hàm tìm max f(a)