$a) \sqrt{4x-1}+\sqrt{4x^2-1}=1$

$b) \sqrt{x^2-x+3}-\sqrt{2+x-x^2}=1$

$c) 4x^3+x-(x+1)\sqrt{2x+1}=0$

$d) \frac{x^2+2x-8}{x^2-2x+3}=(x+1)(\sqrt{x+2}-2)$
Câu d:
Đề thi THPT QG môn Toán 2015:
http://tin.tuyensinh247.com/dap-an-de-thi-thpt-quoc-gia-mon-toan-cua-bo-gddt-2015-c28a23223.html
Tham khảo:
http://diendantoanhoc.net/topic/162093-fracx22x-8x2-2x3x1sqrtx2-2/
uk :)) –  Trần Trân 16-08-16 07:54 PM
ta ghét nhất dẫn links! –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 16-08-16 04:49 PM
b) Đặt $\begin{cases}\sqrt{x^2-x+3}=a \\ \sqrt{2+x-x^2}=b \end{cases}\Rightarrow \begin{cases}a>0; b>0 \\ a+b=5 \end{cases}$
Mặt khác từ pt ta có : $a-b=1$.
kết hợp giải ra ta có: $a=3;b=2\Rightarrow ...............$
c) ĐK:......................
$Pt\Leftrightarrow 8x^3+2x-2(x+1)\sqrt{2x+1}=0$
$\Leftrightarrow 8x^3+2x=(2x+1)\sqrt{2x+1}+\sqrt{2x+1}$
Xét hàm số $f(t)=t^3+t\rightarrow f'(t)=3t^2+1>0\Rightarrow $ Đồng biến
Do đó $Pt\Leftrightarrow 2x=\sqrt{2x+1} $. Bạn tự giải nốt nhé!
a) ĐK:....................
$Pt\Leftrightarrow \sqrt{4x-1}-1+\sqrt{4x^2-1}=0$
$\Leftrightarrow \frac{4x-2}{\sqrt{4x-1}+1}+\sqrt{(2x-1)(2x+1)}=0$
$\Leftrightarrow \sqrt{2x-1}.(\frac{2\sqrt{2x-1}}{\sqrt{4x-1}}+\sqrt{2x+1})=0$
Từ ĐKXĐ: suy ra $x\geq \frac{1}{2}$ Do đó phần trong $(...........)$ vô nghiệm
Vậy $\sqrt{2x-1}=0\Rightarrow ......................$

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