Bài 1:

a) \(3x\left(5x^2-2x+1\right)\)

\(=15x^3-6x^2+3x\)

b) \(\left(x^2-1\right)\left(x^2+2x\right)\)

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

\(=x^4-x^2+2x^3-2x\)

\(=x^4+2x^3-x^2-2x\)

Bài 2:

a) \(3x^2=2x\)

\(\Leftrightarrow3x^2-2x=0\)

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

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

b)\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-2x-1=12x-5\)

\(\Leftrightarrow14x=4\Leftrightarrow x=\frac{2}{7}\)

\(A=\left(2x-1\right)\left(4x+3\right)-\left(2x+5\right)^2-10x\)

\(=\left(8x^2+2x-3\right)-\left(4x^2+20x+25\right)-10x\)

\(=8x^2+2x-3-4x^2-20x-25-10x\)

\(=4x^2-28x-28\)

A=(2x-1)(4x+3)-(2x+5)²-10x

A=8x²+6x-4x-3-(4x²+10x+25)-10x

A=8x²+2x-3-4x²-10x-25-10x

A=4x²-18x-28

Ta có: \(x^2+x+6=y^2\Rightarrow4x^2+4x+24=4y^2\Rightarrow\left(4x^2+4x+1\right)+23=\left(2y\right)^2\)

\(\Rightarrow\left(2x+1\right)^2+23=\left(2y\right)^2\Rightarrow\left(2y\right)^2-\left(2x+1\right)^2=23\)
\(\Rightarrow\left(2y+2x+1\right)\left(2y-2x-1\right)=23\)
Lập bảng:
 

(TM)
Vậy pt có 4 nghiệm nguyên (x;y) \(\in\left\{\left(-6;-6\right),\left(5;-6\right),\left(6;6\right),\left(5;6\right)\right\}\)