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2. ĐK: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x}\ge0\\b=\sqrt{x^2+4}\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=2a^2\\x^2+4=b^2\\3\sqrt{x^3+4x}=3ab\end{matrix}\right.\)
pt trên được viết lại thành
\(2a^2+b^2-3ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=\dfrac{1}{2}b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{x^2+4}\\\sqrt{x}=\dfrac{1}{2}\sqrt{x^2+4}\end{matrix}\right.\)
Đến đây dễ rồi nhé ^^
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)
ĐK:\(x\ge-3\)
\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)
\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)
\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)
\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)
\(4x^2-4-3x=\sqrt[3]{x^2\left(x^2-1\right)}\)
\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)-3x=\sqrt[3]{x^2\left(x-1\right)\left(x+1\right)}\)
dat \(\left(x-1\right)\left(x+1\right)=y\)
\(4y-3x=\sqrt[3]{x^2y}\)
\(\Leftrightarrow\left(4y-3x\right)^3=x^2y\)
\(\Leftrightarrow64y^3-144y^2x+108yx^2-27x^3=x^2y\)
\(\Leftrightarrow64y^3-144y^2x+107yx^2-27x^3=0\)
\(\Leftrightarrow64y^3-64y^2x-80y^2x+80x^2y+27x^2y-27x^3=0\)
\(\Leftrightarrow\left(y-x\right)\left(64y^2-80xy+27x^2\right)=0\)
de thay \(64y^2-80xy+27x^2=\left(8y\right)^2-2.8y.5x+25x^2+2x^2=\left(8y-5x\right)^2+2x^2>0\)
\(\Rightarrow y=x\)hay \(\left(x-1\right)\left(x+1\right)=x\Rightarrow x^2-x-1=0\)
\(\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+1}{2}\\x=\frac{-\sqrt{5}+1}{2}\end{cases}}\)
câu b tương tự nhé bạn