mọi người ơi câu b là giá trị tuyệt đối của x^2 -1 nha

giúp mình mình tick cho

a) \(\Leftrightarrow x^2+\dfrac{2}{3}x-x^2+\dfrac{3}{4}x=\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{17}{12}x=\dfrac{7}{12}\Leftrightarrow x=\dfrac{7}{17}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow x^3=4\sqrt{2}-3x\) \(\Leftrightarrow x^3+3x=4\sqrt{2}\) (1)

Có \(y^3=17+12\sqrt{2}-3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\right)-\left(17-12\sqrt{2}\right)\)

\(\Leftrightarrow y^3=24\sqrt{2}-3y\) \(\Leftrightarrow y^3+3y=24\sqrt{2}\) (2)

Từ (1) (2)\(\Rightarrow x^3+3x-y^3-3y=-20\sqrt{2}\)

Có \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)=\left(x-y\right)\left[\left(x-y\right)^2+3\left(xy+1\right)\right]\)

\(=\left(x-y\right)\left(x^2+xy+y^2+3\right)=x^3-y^3+3\left(x-y\right)=-20\sqrt{2}\)

Vậy \(M=-20\sqrt{2}\)

theo bài ra

\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)

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

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

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

\(x^3=4\sqrt{2}-3.\left[\sqrt[3]{9-\left(2\sqrt{2}\right)^2}\right]x\)

\(x^3=4\sqrt{2}-3.1x\)

\(x^3=4\sqrt{2}-3x\)

\(< =>x^3+3x-4\sqrt{2}=0\)

rồi làm y tương tự rồi thế vào M là ra

Ta có: \(\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^x}=10\)

\(\Leftrightarrow\sqrt{\left(5+2\sqrt{6}\right)^x}=10-5+2\sqrt{6}=5+2\sqrt{6}\)

\(\Leftrightarrow\left(5+2\sqrt{6}\right)^x=\left(5+2\sqrt{6}\right)^2\)

hay x=2

\(B=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)

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

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

Ta có: \(x=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=1\)

Thay x=1 vào B, ta được:

\(B=-\sqrt{1}\cdot\left(\sqrt{1}-1\right)=0\)