Giải các pt sau:

   1> $\sqrt{2x^{2}+8x+6} +\sqrt{x^{2}-1} = 2x+2 $

   2> $2(1-x)\sqrt{x^{2}+2x- 1} = x^{2} -2x-1$

   3> $\frac{2x^{2}}{(3 - \sqrt{9 +2x})^{2}} = x +21$

   4> $(4x-1)\sqrt{x^{2}+1} = 2x^{2}+ 2x+1$

   5> $x^{2}+ \sqrt{x+5} = 5$

   6> $\sqrt[3]{2x-1} = x\sqrt[3]{16} - \sqrt[3]{2x+1}$

   7> $-x^{2}+ 2 = \sqrt{2-x}$

   8> $\sqrt{x^{2}+15} = 3x +2 + \sqrt{x^{2}+8}$
đúng tick V jum nha –  Lionel Messi 28-07-16 03:25 PM
bạn nên đăng lẻ ra, đăng thế này nhiễu quá –  Confusion 28-07-16 10:28 AM
4)$\Leftrightarrow [2(2x-1)+1]\sqrt{1+x^{2}}=2(x^{2}+1)+ (2x-1) $
Đặt $a=2x-1;b=\sqrt{x^2+1}$ta co $(2a+1)b=2b^{2}+a $
$\Leftrightarrow  (2b-1)(a-b)=0 $ giai ra ta dc x=4/3 
em nghi bai nay de ma em ms len 9 thoi anh nen dong nao ty nhe
7)Đặt $a=\sqrt{2-x}\Rightarrow a^2+x-2=0$
Khi đó ta có hệ $\begin{cases}a^2+x-2=0 \\ x^2+a-2=0 \end{cases}$
Giải hệ tìm x
5) Dặt $\sqrt{x+5}=a\Rightarrow a^2-x-5=0$
Khi đó pt thành: $x^2-a-5=0$
ta có hệ $\begin{cases}x^2-a-5=0 \\ a^2-x-5=0 \end{cases}$
Giải hệ rồi tìm x
2) $PT\Rightarrow 4(1-x)^2(x^2+2x-1)=(x^2-2x-1)^2$
$\Leftrightarrow (3x^2-2x-1)(x^2+2x-5)=0$
$\Rightarrow x$
1)$PT\Rightarrow (\sqrt{2x^2+8x+6}+\sqrt{x^2-1})^2=(2x+2)^2$
$\Rightarrow x^2-1=2\sqrt{(2x^2+8x+6)(x^2-1)}$
$\Leftrightarrow (x^2-1)^2=4(2x^2+8x+6)(x^2-1)$
$\Leftrightarrow (x-1)(x+1)(7x^2+32x+25)=0$

7)$\Leftrightarrow-x^{2}+2-1=can.... -1$
nhan lien hop co $(1-x)(1+x +\frac{1}{(can 2-x) +1})=0$
suy ra x=1
ko để ý hả,thế c báo cáo nhá e :)) –  Ngọc HTN 29-07-16 11:45 AM
thiếu nghiệm rồi thanh niên –  ๖ۣۜJinღ๖ۣۜKaido 29-07-16 11:16 AM
ns chung la nhan lien hop e ko de y lam –  kaito bing 29-07-16 10:34 AM
liên hợp sai dấu òi em –  ๖ۣۜJinღ๖ۣۜKaido 28-07-16 02:56 PM

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