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PT
1
DN
30 tháng 7 2015
\(1, \frac{5x^2+1}{8x^2}=\frac{5\left(5x^2+1\right)\left(4x-6\right)}{5.8x^2\left(4x-6\right)}=\frac{5\left(5x^2+1\right)\left(4x-6\right)}{40x^2\left(4x-6\right)}\)
\(\frac{3x}{2x+3x}=\frac{3x}{5x}=\frac{3x.8x\left(4x-6\right)}{8x.5x\left(4x-6\right)}=\frac{24x^2\left(4x-6\right)}{40x^2\left(4x-6\right)}\)
\(\frac{-5}{4x-6}=\frac{-5.5x.8x}{8x.5x.\left(4x-6\right)}=\frac{-200x^2}{40x^2\left(4x-6\right)}\)
18 tháng 11 2023
a: ĐKXĐ: \(x\notin\left\{4;-4\right\}\)
\(\dfrac{7}{4x+16}=\dfrac{7}{4\left(x+4\right)}=\dfrac{7\left(x-4\right)}{4\left(x+4\right)\left(x-4\right)}\)
\(\dfrac{11}{x^2-16}=\dfrac{11\cdot4}{4\left(x^2-16\right)}=\dfrac{44}{4\left(x-4\right)\left(x+4\right)}\)
b: \(\dfrac{6}{x\left(x+3\right)^2};\dfrac{x-3}{2x\left(x+3\right)^2}\)
ĐKXĐ: \(x\notin\left\{0;-3\right\}\)
\(\dfrac{6}{x\left(x+3\right)^2}=\dfrac{6\cdot2}{2x\left(x+3\right)^2}=\dfrac{12}{2x\left(x+3\right)^2}\)
\(\dfrac{x-3}{2x\left(x+3\right)^2}=\dfrac{x-3}{2x\left(x+3\right)^2}\)
c: \(\dfrac{-6}{1-x};\dfrac{3x}{x^2+x+1};\dfrac{x^2-3x+5}{x^3-1}\)
ĐKXĐ: \(x\ne1\)
\(-\dfrac{6}{1-x}=\dfrac{6}{x-1}=\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{3x}{x^2+x+1}=\dfrac{3x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x^2-3x+5}{x^3-1}=\dfrac{x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
d: \(\dfrac{17}{5x};\dfrac{24}{x-2y};\dfrac{x-y}{8y^2-2x^2}\)
ĐKXĐ: \(x\ne0;x\ne\pm2y\)
\(\dfrac{17}{5x}=\dfrac{17\cdot2\left(x-2y\right)\left(x+2y\right)}{5x\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}=\dfrac{34\left(x^2-4y^2\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)
\(\dfrac{24}{x-2y}=\dfrac{24\cdot10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{240x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)
\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{-5x\left(x-y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{-5x^2+5xy}{10x\left(x-2y\right)\left(x+2y\right)}\)
PN
25 tháng 11 2015
\(a.\) Ta có:
\(MTC:\) \(\left(x+1\right)\left(x+2\right)\)
Do đó
\(\frac{3x}{x+1}=\frac{3x\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}\)
\(\frac{x+4}{x+2}=\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x+2\right)}\)
\(b.\) Ta có:
\(x^2+x=x\left(x+1\right)\)
\(x^2-1=\left(x-1\right)\left(x+1\right)\)
nên \(MTC:\) \(x\left(x-1\right)\left(x+1\right)\)
Do đó:
\(\frac{5}{x^2+x}=\frac{5}{x\left(x+1\right)}=\frac{5\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(\frac{6}{x^2-1}=\frac{6}{\left(x-1\right)\left(x+1\right)}=\frac{6x}{x\left(x-1\right)\left(x+1\right)}\)
\(c.\) Ta có:
\(x^2-5x+4=x^2-x-4x+4=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
\(2x^2-8x=2x\left(x-4\right)\)
nên \(MTC:\) \(2x\left(x-1\right)\left(x-4\right)\)
Do đó:
\(\frac{4}{x^2-5x+4}=\frac{4}{\left(x-1\right)\left(x-4\right)}=\frac{8x}{2x\left(x-1\right)\left(x-4\right)}\)
\(\frac{x+1}{2x^2-8x}=\frac{x+1}{2x\left(x-4\right)}=\frac{\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x-4\right)}\)
27 tháng 11 2020
Làm nốt d :P
\(\frac{x+3}{2x^2-15x-8};\frac{3}{x^2-8x}\)
Ta có : \(2x^2-15x-8=\left(2x+1\right)\left(x-8\right)\)
\(x^2-8x=x\left(x-8\right)\)
MTC : \(x\left(x-8\right)\left(2x+1\right)\)
\(\frac{x+3}{2x^2-15x-8}=\frac{x+3}{\left(2x+1\right)\left(x-8\right)}=\frac{x^2+3x}{x\left(x-8\right)\left(2x+1\right)}\)
\(\frac{3}{x^2-8x}=\frac{3}{x\left(x-8\right)}=\frac{6x+3}{x\left(x-8\right)\left(2x+1\right)}\)
11 tháng 10 2023
a: \(\dfrac{3}{x-1}=\dfrac{3\cdot9}{9\cdot\left(x-1\right)}=\dfrac{27}{9\left(x-1\right)}\)
\(\dfrac{4}{3x-3}=\dfrac{12}{9x-9}=\dfrac{12}{9\left(x-1\right)}\)
\(\dfrac{10}{9-9x}=\dfrac{-10}{9x-9}=-\dfrac{10}{9\left(x-1\right)}\)
b: \(\dfrac{3}{2\left(x-3\right)}=\dfrac{3x-9}{2\left(x-3\right)^2}\)
\(\dfrac{3x-2}{x^2-6x+9}=\dfrac{6x-4}{2\left(x-3\right)^2}\)
c: \(\dfrac{3}{x^2+2x+1}=\dfrac{3}{\left(x+1\right)^2}=\dfrac{3x}{x\left(x+1\right)^2}\)
\(-\dfrac{2}{x^2+x}=\dfrac{-2}{x\left(x+1\right)}=\dfrac{-2\left(x+1\right)}{x\left(x+1\right)^2}\)
