QUy đồng
\(\dfrac{5x^2}{x^2+5x+6};\dfrac{2x+3}{x^2+7x+10};-5\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
BN
1
26 tháng 5 2022
\(\dfrac{5x^2}{x^2+5x+6}=\dfrac{5x^2}{\left(x+2\right)\left(x+3\right)}=\dfrac{5x^2\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)
\(\dfrac{2x+3}{x^2+7x+10}=\dfrac{2x+3}{\left(x+2\right)\left(x+5\right)}=\dfrac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)
\(-5=\dfrac{-5\left(x+2\right)\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)
8 tháng 12 2021
\(a,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ b,=\dfrac{x^2-x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+x-6}{x^2+x-6}=1\)
8 tháng 12 2021
\(a,\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{25-x}{5\left(5-x\right)}\)
\(=\dfrac{-3x-5}{x\left(5-x\right)}+\dfrac{25-x}{5\left(5-x\right)}\)
\(=\dfrac{5\left(-3x-5\right)}{5x\left(5-x\right)}+\dfrac{x\left(25-x\right)}{5x\left(5-x\right)}\)
\(=\dfrac{-15x-25+25x-x^2}{5x\left(5-x\right)}\)
\(=\dfrac{10x-25-x^2}{5x\left(5-x\right)}\)
\(=\dfrac{-\left(5-x\right)^2}{5x\left(5-x\right)}\)
\(=\dfrac{-5+x}{5x}\)
\(b,\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)
\(=\dfrac{x+1}{x+3}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-2x+x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2+x-6}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2+x-6}{x^2-2x+3x-6}\)
\(=\dfrac{x^2+x-6}{x^2+x-6}\)
\(=1\)
NN
19 tháng 11 2017
a)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}=\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\)
MTC: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\ =\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x-5\right)\left(5x-5\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
qui đồng cái nào hả bn ?? ?
MSC:
\(\dfrac{5x^2}{\left(x+2\right)\left(x+3\right)};\dfrac{\left(2x+3\right)}{\left(x+2\right)\left(x+5\right)};-5\)
\(\dfrac{5x^2\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)};\dfrac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)};\dfrac{-5\left(x+2\right)\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)