Bài 4:
a) Ta có: \(A=2\left(x-3\right)+5\left(x+2\right)-7x+2\)
\(=2x-6+5x+10-7x+2\)
\(=6\)
b) Ta có: \(B=2\left(3x-3\right)-3\left(x+5\right)-3x\)
\(=6x-6-3x-15-3x\)
=-21
c) Ta có: \(C=\left(x+2\right)\left(x-3\right)-\left(x+3\right)\left(x-4\right)\)
\(=x^2-x-6-x^2+x+12\)
=6
d) Ta có: \(D=\left(3x+1\right)\left(2x+3\right)-x\left(6x+11\right)-7\)
\(=6x^2+9x+2x+3-6x^2-11x-7\)
=-4
e) Ta có: \(D=\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3-1-x^3-1\)
=-2
Bài 3:
a) Ta có: \(\left(2x+5\right)\left(3x-9\right)-6x^2=20\)
\(\Leftrightarrow6x^2-18x+15x-45-6x^2=20\)
\(\Leftrightarrow-3x=65\)
hay \(x=-\dfrac{65}{3}\)
b) Ta có: \(\left(2x-1\right)\left(3x+1\right)-\left(3x-4\right)\left(2x+3\right)=5\)
\(\Leftrightarrow6x^2+2x-3x-1-6x^2-9x+8x+12=5\)
\(\Leftrightarrow-2x=-6\)
hay x=3
d) Ta có: \(\left(x-3\right)\left(x+5\right)=\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+5x-3x-15=x^2-2x+3x-6\)
\(\Leftrightarrow x^2+2x-15-x^2-x+6=0\)
\(\Leftrightarrow x-9=0\)
hay x=9
c) Ta có: \(\left(9-4x^2\right)\left(9+4x^2\right)+16x^4-1=2x\)
\(\Leftrightarrow81-16x^4+16x^4-1=2x\)
\(\Leftrightarrow2x=80\)
hay x=40
e) Ta có: \(\left(3x-7\right)\left(2x-3\right)=6x^2\)
\(\Leftrightarrow6x^2-9x-14x+21-6x^2=0\)
\(\Leftrightarrow-23x=-21\)
hay \(x=\dfrac{21}{23}\)
f) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x^2-3\right)=5\)
\(\Leftrightarrow x^3-8-x^3+3x=5\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
