QUY ĐỒNG MẪU CÁC PHÂN THỨC SAU:
b) \(\frac{5}{2x+6}và\frac{3}{x^2-9}\)
c) \(\frac{3}{x^2-5x}và\frac{-5}{10-2x}\)
d) \(\frac{1-3x}{2x};\frac{2}{x-3};\frac{5x-6}{9-x^2}\)
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16 tháng 12 2019
b) \(\hept{\begin{cases}\frac{5}{2x+6}=\frac{5}{2\left(x+3\right)}\\\frac{3}{x^2-9}=\frac{3}{\left(x+3\right)\left(x-3\right)}\end{cases}}\)
\(\Rightarrow MTC=2\left(x+3\right)\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{5}{2\left(x+3\right)}=\frac{5\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}\\\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{6}{2\left(x-2\right)\left(x+3\right)}\end{cases}}\)
CÒn lại tương tự nhé !
14 tháng 2 2020
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
27 tháng 11 2020
a, \(\frac{3x}{2x+4};\frac{x+3}{x^2-4}\)
Ta có : \(2x+4=2\left(x+2\right)\)
\(x^2-4=\left(x-2\right)\left(x+2\right)\)
MTC : \(2\left(x-2\right)\left(x+2\right)\)
\(\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}=\frac{3x\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\frac{3x^2-6x}{2\left(x-2\right)\left(x+2\right)}\)
\(\frac{x+3}{x^2-4}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}=\frac{2x+6}{\left(x-2\right)\left(x+2\right)}\)
27 tháng 11 2020
c, \(\frac{2x}{x^2-8x+16};\frac{x}{3x^2-12x}\)
Ta có : \(x^2-8x+16=\left(x-4\right)^2\)
\(3x^2-12x=3x\left(x-4\right)\)
MTC : \(3x\left(x-4\right)^2\)
\(\frac{2x}{x^2-8x+16}=\frac{2x}{\left(x-4\right)^2}=\frac{6x^2}{3x\left(x-4\right)^2}\)
\(\frac{x}{3x^2-12x}=\frac{x}{3x\left(x-4\right)}=\frac{x^2+4x}{3x\left(x-4\right)\left(x+4\right)}\)