Bài 22:
1: \(3a^3\left(x^2-1\right)^4:\left\lbrack3a^3\left(x^2-1\right)^3\right\rbrack=15\)
=>\(\frac{3a^3}{3a^3}\cdot\frac{\left(x^2-1\right)^4}{\left(x^2-1\right)^3}=15\)
=>\(x^2-1=15\)
=>\(x^2=16\)
=>x=4 hoặc x=-4
2: \(x^3\left(2x-1\right)^{m+2}:\left\lbrack x^3\left(2x-1\right)^{m-1}\right\rbrack-3^5:3^2=0\)
=>\(\left(2x-1\right)^{m+2-m+1}=3^3\)
=>\(\left(2x-1\right)^3=3^3\)
=>2x-1=3
=>2x=4
=>x=2
3: \(\left(3x^5-4x^3\right):x^3-\left(3x+1\right)^2:\left(3x+1\right)-3x^7:x^5=0\)
=>\(3x^2-4-\left(3x+1\right)-3x^2=0\)
=>-4-3x-1=0
=>3x+5=0
=>3x=-5
=>x=-5/3
Bài 23:
1:
\(3x^5-x^4-2x^3+x^2+4x+5\)
\(=3x^5-6x^4+6x^3+5x^4-10x^3+10x^2+2x^3-4x^2+4x-5x^2+10x-10-10x+15\)
\(=\left(x^2-2x+2\right)\left(3x^3+5x^2+2x-5\right)-10x+15\)
=>\(\frac{3x^5-x^4-2x^3+x^2+4x+5}{x^2-2x+2}=3x^3+5x^2+2x-5\) dư -10x+15
Để dư bằng 0 thì -10x+15=0
=>-10x=-15
=>x=1,5
2:
\(\frac{x^5+2x^4+3x^2+x-3}{x^2+1}=\frac{x^5+x^3+2x^4+2x^2-x^3-x+x^2+1+2x-4}{x^2+1}\)
\(=x^3+2x^2-x+1+\frac{2x-4}{x^2+1}\)
Để dư bằng 0 thì 2x-4=0
=>2x=4
=>x=2
