B: rút gọn

a) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-6x^2+12x\)

\(=x^3-6x^2+12x-8\)

\(=\left(x-2\right)^3\)

b) Ta có: \(\left(2x+5\right)\left(5-2x\right)+\left(x-5\right)\left(4x+5\right)\)

\(=25-4x^2+4x^2+5x-20x-25\)

=-15x

\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)

\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)

\(=x^2-2x-5\)

\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)

\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)

\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)

\(=2x-3\)

\(\frac{\left(8x^3-y^3\right)\left(4x^2-y^2\right)}{\left(2x+y\right)\left(4x^2-4xy+y^2\right)}\)

\(=\frac{\left(2x-y\right)\left(4x^2+2xy+y^2\right)\left(2x-y\right)\left(2x+y\right)}{\left(2x+y\right)\left(2x-y\right)^2}\)

\(=4x^2+2xy+y^2\)

a: \(y=\left(x+2\right)\left(2x^2-3\right)\)

=>\(y'=\left(x+2\right)'\left(2x^2-3\right)+\left(x+2\right)\left(2x^2-3\right)'\)

=>\(y'=2x^2-3+\left(x+2\right)\cdot2x\)

\(\Leftrightarrow y'=2x^2-3+2x^2+4x=4x^2+4x-3\)

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

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

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

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

=>\(y'=2x^2+4x-2x-4+x^2-2x+1\)

=>\(y'=3x^2-3\)

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

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

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

=>\(y'=4x^2+2x+2x^2-2=6x^2+2x-2\)

d: \(y=\left(x+2\right)\left(2x^2-5\right)\)

=>\(y'=\left(x+2\right)'\left(2x^2-5\right)+\left(x+2\right)\left(2x^2-5\right)'\)

=>\(y'=2x^2-5+2x\left(x+2\right)=4x^2+4x-5\)

a) Ta có: \(5x^2-3x\left(x+2\right)\)

\(=5x^2-3x^2-6x\)

\(=2x^2-6x\)

b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)

\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)

\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)

\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)

d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)

\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)

\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)

\(=-4x^2y+5x^2-2x\)

e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)

\(=4x^4-16x^3+4x^4-2x^3+14x^2\)

\(=8x^4-18x^3+14x^2\)

f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)

\(=25x-12x+4+35x-14x^3\)

\(=-14x^3+48x+4\)

a: \(y=\left(x-1\right)^3\)

=>\(y'=\left[\left(x-1\right)^3\right]'=3\left(x-1\right)^2\cdot\left(x-1\right)'\)

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

b: \(y=\left(x+2\right)\left(2x^2-3\right)\)

=>\(y'=\left(x+2\right)'\left(2x^2-3\right)+\left(x+2\right)\left(2x^2-3\right)'\)

=>\(y'=2x^2-3+2\left(x+2\right)\)

\(=2x^2+2x+1\)

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

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

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

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

\(=2x^2+4x-2x-4-x^2+2x-1\)

=>\(y'=x^2+4x-5\)

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

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

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

\(=4x^2+2x+2x^2-2=6x^2+2x-2\)

Đặt \(y = 2x - 5\).

 \(\begin{array}{l}\left[ {8{x^3}{{\left( {2x - 5} \right)}^2} - 6{x^2}{{\left( {2x - 5} \right)}^3} + 10x{{\left( {2x - 5} \right)}^2}} \right]:2x{\left( {2x - 5} \right)^2}\\ = \left( {8{x^3}.{y^2} - 6{x^2}.{y^3} + 10x.{y^2}} \right):2x{y^2}\\ = 8{x^3}.{y^2}:2x{y^2} - 6{x^2}.{y^3}:2x{y^2} + 10x.{y^2}:2x{y^2}\\ = 4{x^2} - 3xy + 5\\ = 4{x^2} - 3x\left( {2x - 5} \right) + 5\\ = 4{x^2} - 6{x^2} + 15x + 5\\ =  - 2{x^2} + 15x + 5\end{array}\) 

a/

\(\Leftrightarrow4x^2-12x+9=\left(3x-2\right)^2\)

\(\Leftrightarrow5x^2-5=0\Rightarrow x=\pm1\)

b/

\(\Leftrightarrow25x^2-10x+1=\left(x+6\right)^2\)

\(\Leftrightarrow24x^2-22x-35=0\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=-\frac{5}{6}\end{matrix}\right.\)

c/

\(\Leftrightarrow16x^2-8x+1=\left(x-3\right)^2\)

\(\Leftrightarrow15x^2-2x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{2}{3}\end{matrix}\right.\)

d/ \(x\ge\frac{3}{2}\)

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

\(\Leftrightarrow21x^2+22x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{2}{7}\\x=-\frac{4}{3}\end{matrix}\right.\)

e/

\(\Leftrightarrow\left[{}\begin{matrix}3x-4=x-2\\3x-4=2-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\4x=6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)

f/

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)

g/

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\3x^2-3x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\frac{3\pm\sqrt{33}}{6}\\\end{matrix}\right.\)