ĐKXĐ:....

\(\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}\)

\(\Rightarrow4-\sqrt{1-x}=2-x\)

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

\(\Rightarrow1-x=4+4x+x^2\)

\(\Rightarrow1-x-4-4-x^2=0\)

\(\Rightarrow x^2+x+7=0\)

Đến đây dễ rồi làm nốt nha bạn !

 ĐKXĐ:....

\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}4−1−x​​=2−x​

\Rightarrow4-\sqrt{1-x}=2-x⇒4−1−x​=2−x

\Rightarrow\sqrt{1-x}=2+x⇒1−x​=2+x

\Rightarrow1-x=4+4x+x^2⇒1−x=4+4x+x2

\Rightarrow1-x-4-4-x^2=0⇒1−x−4−4−x2=0

\Rightarrow x^2+x+7=0⇒x2+x+7=0

Đến đây dễ rồi làm nốt nha bạn !

\(DK:x\ge-1\)

\(\Leftrightarrow\left(\sqrt[3]{x-2}-1\right)+\left(\sqrt{x+1}-2\right)=0\)

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

\(\Leftrightarrow\left(x-3\right)\left[\frac{1}{\left(\sqrt[3]{x-2}\right)+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}\right]=0\)

Vi \(\frac{1}{\left(\sqrt[3]{x-2}\right)^2+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}>0\)

\(\Rightarrow x=3\left(n\right)\)

Vay PT co nhiem la \(x=3\)

\(DK:x\ge\sqrt[3]{\frac{2}{3}}\)

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

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

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

Vi PT trong cai ngoac thu 2 >0

\(\Rightarrow x=1\left(n\right)\)

Vay nghiem cua PT la \(x=1\)

\(A=\frac{\sqrt{2\left(x+1\right)}}{\sqrt{2}}+\frac{\sqrt{2\left(y+1\right)}}{\sqrt{2}}+\frac{\sqrt{2\left(z+1\right)}}{\sqrt{2}}\le\frac{x+y+z+9}{2\sqrt{2}}=3\sqrt{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\)