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

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

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

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

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

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

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

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

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

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

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

\(a^3-x-x^3+a=0\)

\(\Leftrightarrow a^3-x^3+\left(a-x\right)=0\)

\(\Leftrightarrow\left(a-x\right)\left(a^2+ax+x^2\right)+\left(a-x\right)=0\)

\(\Leftrightarrow\left(a-x\right)\left(a^2+ax+x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=x\\a^2+ax+x^2+1=0\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow a^2+ax+x^2+1=0\)

Ta có:\(a^2+ax+x^2+1=a^2+2.a.\dfrac{1}{2}x+\dfrac{1}{4}x^2+\dfrac{3}{4}y^2+1\)

                                     \(=\left(a+\dfrac{1}{2}x\right)^2+\dfrac{3}{4}y^2+1>0\)

\(\Rightarrow\left(2\right)\) vô lý

Vậy \(a=x\)

`@`\(189+424+511+276+55\)

\(=\left(189+511\right)+\left(424+276\right)+55\)

\(=700+700+55\)

\(=1455\)

`@`\(36\times28+36\times81+64\times141-64\times41\)

\(=36\times\left(28+82\right)+64\times\left(141-41\right)\)

\(=36\times110+64\times100\)

\(=3960+6400\)

\(=10360\)

`@`\(5+8+11+14+...+302\)

\(=\left(302+5\right)\times100:2\)

\(=15350\)

`a)`\(A=\dfrac{x}{x-1}+\dfrac{1}{x+2}-\dfrac{3x}{x^2+x-2}\)

\(A=\dfrac{x}{x-1}+\dfrac{1}{x+2}-\dfrac{3x}{\left(x-1\right)\left(x+2\right)}\)

\(A=\dfrac{x\left(x+2\right)+\left(x-1\right)-3x}{\left(x-1\right)\left(x+2\right)}\)

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

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

\(A=\dfrac{x+1}{x+2}\)

`b)`\(S=A.B\)

\(S=\dfrac{x+1}{x+2}.\dfrac{x+3}{x+1}\)

\(S=\dfrac{x+3}{x+2}\)

\(S=\dfrac{x+2+1}{x+2}=1+\dfrac{1}{x+2}\)

Ta có:\(x\ge0\Rightarrow x+2\ge2\)

\(\Rightarrow S\le1+\dfrac{1}{2}=\dfrac{3}{2}\)

Vậy \(Max_S=\dfrac{3}{2}\) khi \(x=0\)