Rút gọn phân thức sau
\(\frac{2x^2-3x-20}{x^2-16}\)
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31 tháng 10 2016
Câu trả lời là:
\(\frac{3x+6}{2x^2+x-6}=\frac{3x+6}{5x-6}=2x\)
Đ/s: 2x
đúng cho một tích nha
QA
21 tháng 7 2021
Trả lời:
a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)
b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)
c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)
d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)
e, \(E=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
27 tháng 12 2015
a,\(\frac{2x^2+4x}{x+2}\)=\(\frac{2x\left(x+2\right)}{x+2}\)\(=2x\)
b, \(\frac{3x}{2x+4}\)=\(\frac{3x^2-6x}{2\left(x+2\right)\left(x-2\right)}\)
\(\frac{x+3}{x^2+4}\)=\(\frac{2x+6}{2\left(x-2\right)\left(x+2\right)}\)
tick mình nhé!!
A
25 tháng 12 2018
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(\frac{2x^2-3x-20}{x^2-16}\)
\(=\frac{2x^2-8x+5x-20}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{2x\left(x-4\right)+5\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\)
\(=\frac{\left(2x+5\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)
\(=\frac{2x+5}{x+4}\)
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