x^2(x-5)-4(x-5)=0

=> (x^2 - 4)(x-5) = 0

=> (x-2)(x+2)(x-5) = 0

=> x - 2 = 0 hoặc x + 2 = 0 hoặc x - 5 = 0

=> x = 2 hoặc x = -2 hoặc x = 5

\(x^2\left(x-5\right)-4\left(x-5\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left(x-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+2\right)\left(x-5\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2=0\\x+2=0\\x-5=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x=2\\x=-2\\x=5\end{cases}}\)

\(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow9x^2-1=15x^2+24x-5x-8\)

\(\Leftrightarrow9x^2-1=15x^2+19x-8\)

\(\Leftrightarrow9x^2-1-15x^2-19x+8=0\)

\(\Leftrightarrow-6x^2+7-19x=0\)

\(\Leftrightarrow6x^2+19x-7=0\)

\(\Leftrightarrow6x^2+21x-2x-7=0\)

\(\Leftrightarrow3x\left(2x+7\right)-\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x+7\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\3x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=\frac{1}{3}\end{cases}}\)

Vậy: Phương trình có tập nghiệm là: S = {-7/2; 1/3}

\(B=-4x^2+x\)\(=-\left(4x^2-x\right)\)

\(=-\left[\left(2x\right)^2-2\cdot2x\cdot\frac{1}{4}+\left(\frac{1}{4}\right)^2\right]+\left(\frac{1}{4}\right)^2\)

\(=-\left(2x-\frac{1}{4}\right)^2+\frac{1}{16}\)

Vì : \(\left(2x-\frac{1}{4}\right)^2\ge0\forall x\in R\)

\(\Rightarrow-\left(2x-\frac{1}{4}\right)^2\le0\forall x\in R\)

\(\Rightarrow B\le\frac{1}{16}\forall x\in R\)

Vậy B có GTLN khi B = \(\frac{1}{16}\)

\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)

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

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

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

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

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

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