a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{-\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x\sqrt{x}+x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)

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

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

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

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

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

b: Thay \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) vào A, ta được:

\(A=\dfrac{17-12\sqrt{2}-\sqrt{2}+1+1}{3-2\sqrt{2}}=\dfrac{19-13\sqrt{2}}{3-2\sqrt{2}}=5-\sqrt{2}\)

b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)

\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)

\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)

\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)

Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)

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

\(A^3=x^3-3x+3A\sqrt[3]{\frac{x^6-6x^4+9x^2-\left(x^6-6x^4+9x^2-4\right)}{4}}\)

\(A^3=x^3-3x+3A\)

\(A^3-x^3-3\left(A-x\right)=0\)

\(\left(A-x\right)\left(A^2+x^2+Ax-3\right)=0\)

\(\Rightarrow A=x\) (do \(\left\{{}\begin{matrix}A>0\\x\ge2\end{matrix}\right.\) \(\Rightarrow x^2-3>0\Rightarrow A^2+x^2+Ax-3>0\))

2/ \(a+1=\sqrt{17}\Rightarrow a^2+2a+1=17\Rightarrow a^2+2a-17=-1\)

\(P=\left[a^3\left(a^2+2a-17\right)-a^2+18a-17\right]^{2018}\)

\(=\left(-a^3-a^2+18a-17\right)^{2018}\)

\(=\left(-a\left(a^2+2a-17\right)+a^2+a-17\right)^{2018}\)

\(=\left(a^2+2a-17\right)^{2018}\)

\(=\left(-1\right)^{2018}=1\)

Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)

\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)

\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)

\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)

\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)

Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình

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