Bài 1:
a: Đặt \(A=5x\left(x^2-4x+1\right)-\left(x^2+1\right)\left(5x-1\right)+20\left(1-x\right)\left(1+x\right)\)
\(=5x^3-20x^2+5x-\left(5x^3-x^2+5x-1\right)+20\left(1-x^2\right)\)
\(=5x^3-20x^2+5x-5x^3+x^2-5x+1+20-20x^2\)
\(=-39x^2+21\)
KHi x=-1 thì \(A=-39\cdot\left(-1\right)^2+21=-39+21=-18\)
b: Đặt \(B=\left(2-x\right)\left(x^2+2x+4\right)+x\left(x^2-1\right)+\frac12\left(8x-3\right)\)
\(=-\left(x-2\right)\left(x^2+2x+4\right)+x^3-x+4x-1,5\)
\(=-\left(x^3-8\right)+x^3+3x-1,5\)
\(=-x^3+8+x^3+3x-1,5=3x+6,5\)
Khi x=1/6 thì \(B=3\cdot\frac16+6,5=0,5+6,5=7\)
Bài 2:
a: \(\left(2x-3\right)\left(x-2\right)-x\left(2x+1\right)-6=0\)
=>\(2x^2-4x-3x+6-2x^2-x-6=0\)
=>-8x=0
=>x=0
b: \(5\left(2x-1\right)\left(x+2\right)-2\left(5x-3\right)\left(x+2\right)=5\)
=>\(5\left(2x^2+4x-x-2\right)-2\left(5x^2+2x-3x-6\right)=5\)
=>\(5\left(2x^2+3x-2\right)-2\left(5x^2-x-6\right)=5\)
=>\(10x^2+15x-10-10x^2+2x+12=5\)
=>17x+2=5
=>17x=3
=>\(x=\frac{3}{17}\)
c: \(2x^2+3\left(x-2\right)\left(x+2\right)-5x\left(x+3\right)=0\)
=>\(2x^2+3\left(x^2-4\right)-5x^2-15x=0\)
=>\(-3x^2-15x+3x^2-12=0\)
=>-15x-12=0
=>15x=-12
=>\(x=-\frac45\)
