ĐK: \(0\le x\le1\)

Đặt \(t=\sqrt{x}+\sqrt{1-x}\) ( \(t>0\) )

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

\(\Leftrightarrow t^2-1=2\sqrt{x-x^2}\)

\(\Leftrightarrow\frac{t^2-1}{2}=\sqrt{x-x^2}\)

Ta có \(pt\Leftrightarrow1+\frac{2}{3}\cdot\frac{t^2-1}{2}=t\)

\(\Leftrightarrow1+\frac{t^2-1}{3}-t=0\)

\(\Leftrightarrow t^2-1-3t+3=0\)

\(\Leftrightarrow t^2-3t+2=0\)

\(\Leftrightarrow\left(t-1\right)\left(t-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

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

TH1: \(\sqrt{x}+\sqrt{1-x}=1\)

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

\(\Leftrightarrow\sqrt{x\left(1-x\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)( thỏa (

TH2: \(\sqrt{x}+\sqrt{1-x}=2\)

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

\(\Leftrightarrow\sqrt{x\left(1-x\right)}=\frac{3}{2}\)

\(\Leftrightarrow x\left(1-x\right)=\frac{9}{4}\)

\(\Leftrightarrow4x\left(1-x\right)=9\)

\(\Leftrightarrow4x^2-4x+9=0\)

\(\Leftrightarrow\left(2x+1\right)^2+8=0\)( vô lý )

Vậy \(x\in\left\{0;1\right\}\)

Dk 1<x<2

√x^2 -x -2<x+2

5x+6>0

X > -6/5

Bpt vô nghiệm

\(A=\left(\frac{\sqrt{3}}{x^2+x\sqrt{x}+3}+\frac{3}{x^3-\sqrt{27}}\right)\left(\frac{x}{\sqrt{3}}+\frac{\sqrt{3}}{x}+1\right)\)

\(\Leftrightarrow A=\left[\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}+\frac{3}{\left(x-\sqrt{3}\right)\left(x+x\sqrt{3}+3\right)}\right]\left(\frac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\right)\)

\(\Leftrightarrow A=\frac{x\sqrt{3}-3+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}.\frac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)

\(\Leftrightarrow A=\frac{1}{x-\sqrt{3}}\)

olm còn lỗi nên ko trình bày bth đc, bn tự viết lại nhá :)) 

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

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

\(\frac{1}{\sqrt{x+1}+\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x+1}+\sqrt{x}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}\)

\(VT=\sqrt{x+3}-\sqrt{x+2}+\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+1}-\sqrt{x}\)

\(VT=\sqrt{x+3}-\sqrt{x}=1\)

Dễ r -,-