fix lai chut...

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Ta có : \(a=2b\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-2x+4=4x+8\)

\(\Leftrightarrow x^2-6x-4=0\)

\(\Delta=6^2-4\cdot\left(-4\right)=52\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{6+\sqrt{52}}{2}=3+\sqrt{13}\\x=\frac{6-\sqrt{52}}{2}=3-\sqrt{13}\end{matrix}\right.\)

Vậy....

ĐK: \(x\ge-2\)

\(2\left(x^2+2\right)=3\left(\sqrt{x^3+8}+2x\right)\)

\(\Leftrightarrow2x^2+4=3\sqrt{x^3+8}+6x\)

\(\Leftrightarrow2x^2-6x+4=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow2\left(x^2-3x+2\right)=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x+4}=a\\\sqrt{x+2}=b\end{matrix}\right.\)( \(a,b\ge0\) )

Ta có : \(a^2-b^2=x^2-2x+4-x-2=x^2-3x+2\)

\(pt\Leftrightarrow2\left(a^2-b^2\right)=3ab\)

\(\Leftrightarrow2a^2-3ab-2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\left(chon\right)\\2a=-b\left(loai\right)\end{matrix}\right.\)

Ta có \(a=2b\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-4x+4=4x+8\)

\(\Leftrightarrow x^2-8x-4=0\)

\(\Delta=8^2-4\cdot\left(-4\right)=80\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{8+\sqrt{80}}{2}\\x=\frac{8-\sqrt{80}}{2}\end{matrix}\right.\)( thỏa )

Vậy...

\(\frac{\left(x+4\right)\left(x-2\right)}{x^2-2x+3}=\left(x+1\right)\frac{x+2-4}{\sqrt{x+2}+2}\)

\(\left(x-2\right)\left(\frac{x+4}{x^2-2x+3}-\frac{x+1}{\sqrt{x+2}+2}\right)=0\)

+ x=2

+ chiu kho lam cai con lai

Lời giải:
ĐK: \(-2\leq x\leq 4\)

Ta có: \(x^2-2x+8-4\sqrt{(4-x)(x+2)}=0\)

\(\Leftrightarrow x^2-2x+8-4\sqrt{2x+8-x^2}=0\)

\(\Leftrightarrow 16-(2x-x^2+8)-4\sqrt{2x+8-x^2}=0\)

Đặt \(\sqrt{2x+8-x^2}=t\)

\(\Rightarrow 16-t^2-4t=0\)

\(\Rightarrow t=-2\pm 2\sqrt{5}\). Vì \(t\geq 0\Rightarrow t=-2+2\sqrt{5}\)

\(\Rightarrow t^2=2x+8-x^2=24-8\sqrt{5}\)

\(\Leftrightarrow x^2-2x+16-8\sqrt{5}=0\)

\(\Rightarrow x=1\pm \sqrt{8\sqrt{5}-15}\) (đều thỏa mãn)

Vậy............

mong mọi người giải giúp em vs