1)$x^{2}-x-4=2\sqrt{x-1}\left ( 1-x \right )$
2) $2x+3+\sqrt{4x^{2}+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}$
1)$x^{2}-x-4=2\sqrt{x-1}\left ( 1-x \right )$
2) $2x+3+\sqrt{4x^{2}+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}$
Câu 1 $x^{2}-x-4=2\sqrt{x-1}(1-x)$ (1) (ĐKXĐ: x$\geq$1)
Đặt a=$\sqrt{x-1}$ , b=1-x (a dương)
khi đó (1)$a^{2}+b^{2}-4=2ab\Leftrightarrow a-b=\pm 2$
TH1 a=b+2$\Leftrightarrow \sqrt{x-1}=3-x \Leftrightarrow \left\{\begin{matrix}x^{2}-7x+10=0 & \\ 1\leq x\leq 3 & \end{matrix}\right.\Leftrightarrow x=2$
Knowing both victory and defeat.That is the way you become a real man-Shanks
2) $2x+3+\sqrt{4x^{2}+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}$
Câu 2 mình nghĩ ở chỗ 2x+3 là dấu trừ
Knowing both victory and defeat.That is the way you become a real man-Shanks
Nếu là dấu trừ thì ta có lời giải sau
Đặt a=$\sqrt{x+2}$ và b=$\sqrt{4x+1}$ với a,b dương
khi đó pt$\Leftrightarrow b^{2}-2a^{2}+ab=2a+b\Leftrightarrow (2a+b)(b-a-1)=0$
Knowing both victory and defeat.That is the way you become a real man-Shanks
1)$x^{2}-x-4=2\sqrt{x-1}\left ( 1-x \right )$
đk: $x\geq 1$
$x^{2}-x-4+2\sqrt{x-1}^{3}=0$
$\Leftrightarrow x^{2}-x-2+2(\sqrt{x-1}^{3}-1)=0$
$(x-2)(x+1)+2.\frac{x-2}{1+\sqrt{x-1}+\sqrt{x-1}^{2}}=0$
$\Leftrightarrow (x-2)(x+1+\frac{2}{1+\sqrt{x+1}+\sqrt{x+1}^{2}})=0\Leftrightarrow x=2$
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