quy đồng các phân thức sau
a,\(\dfrac{x+1}{x-1};\dfrac{x-1}{x+1};\dfrac{4}{1-x^2}\)
b,\(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3};\dfrac{x}{y^2xy}\)
c,\(\dfrac{4x}{x-2};\dfrac{3x}{x-2};\dfrac{12x}{x^2-4}\)
d,\(\dfrac{7}{x};\dfrac{x}{x+6};\dfrac{36}{x^2+6x}\)
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6:
a: ĐKXĐ: x<>0
\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)
\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)
b: ĐKXĐ: x<>1
\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
d: ĐKXĐ: x<>-2
\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)
e: ĐKXĐ: x<>-y
\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)
7:
a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)
\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)
b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)
\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)
c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)
d:
\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)
\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)
a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)
3/x^2-9=6/2(x+3)(x-3)
b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)
\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)
c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)
x/x-y=x^2/x(x-y)
e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
Bài giải
a) \(\dfrac{1}{x+2}=\dfrac{x.\left(x-2\right)}{\left(x+2\right)\left(x-2\right).x}=\dfrac{x^2-2x}{x\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
b) \(x^2+1=\dfrac{x^2+1}{1}=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x^2-1}=\dfrac{x^4-1}{x^2-1}\)
\(\dfrac{x^4}{x^2-1}\) giữ nguyên.
c) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
a: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(2-x\right)\left(2+x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=\dfrac{8x+16}{x\left(2-x\right)\left(2+x\right)}\)
b: \(x^2+1=\dfrac{x^4-1}{x^2-1}\)
\(\dfrac{x^4}{x^2-1}=\dfrac{x^4}{x^2-1}\)
c: \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3y}{y\cdot\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y\left(y-x\right)}=\dfrac{-x}{y\left(x-y\right)}=\dfrac{-x\left(x-y\right)^2}{y\left(x-y\right)^3}\)
\(a,\left(1\right)=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(2\right)=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(3\right)=\dfrac{-4}{\left(x-1\right)\left(x+1\right)}\\ b,\left(1\right)=\dfrac{x^4y^3}{xy^3\left(x-y\right)^3};\left(2\right)=\dfrac{x\left(x-y\right)^3}{xy^3\left(x-y\right)^3}\\ c,\left(1\right)=\dfrac{4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(2\right)=\dfrac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(3\right)=\dfrac{12x}{\left(x-2\right)\left(x+2\right)}\\ d,\left(1\right)=\dfrac{7\left(x+6\right)}{x\left(x+6\right)};\left(2\right)=\dfrac{x^2}{x\left(x+6\right)};\left(3\right)=\dfrac{36}{x\left(x+6\right)}\)