Question

[✉lãɴH⃣╰❥häńZ͜͡╰❥k̫ınżζ●]

giải phương trình sau: 

a)\(2\left(1-x\right)\sqrt{x^2+2x-1}+2x+1=x^2\)

b)\(\sqrt{5x-1}+\sqrt[3]{9-x}=2x^2+3x-1\)

Answer 1

[Nguyễn Việt Lâm]

a.

ĐKXĐ: \(x^2+2x-1\ge0\)

\(x^2+2x-1+2\left(x-1\right)\sqrt{x^2+2x-1}-4x=0\)

Đặt \(\sqrt{x^2+2x-1}=t\ge0\)

\(\Rightarrow t^2+2\left(x-1\right)t-4x=0\)

\(\Delta'=\left(x-1\right)^2+4x=\left(x+1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=1-x+x+1=2\\t=1-x-x-1=-2x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-5=0\\3x^2-2x+1=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=-1\pm\sqrt{6}\)

Answer 2

[Nguyễn Việt Lâm]

b.

ĐKXĐ: \(x\ge\dfrac{1}{5}\)

\(2x^2+x-3+2x-\sqrt{5x-1}+2-\sqrt[3]{9-x}=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\dfrac{\left(x-1\right)\left(4x-1\right)}{2x+\sqrt[]{5x-1}}+\dfrac{x-1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3+\dfrac{4x-1}{2x+\sqrt[]{5x-1}}+\dfrac{1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}\right)=0\)

\(\Leftrightarrow x=1\) (ngoặc đằng sau luôn dương)