Quy đồng mẫu thức các phân thức sau 1/3x+xy, 2y+2x và 1/x^2+2xy+y^2
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Những câu hỏi liên quan
28 tháng 7 2023
a: 1/x^2y=1/x^2y
3/xy=3x/x^2y
b: \(\dfrac{x}{x^2+2xy+y^2}=\dfrac{x}{\left(x+y\right)^2}\)
\(\dfrac{2x}{x^2+xy}=\dfrac{2}{x+y}=\dfrac{2x+2y}{\left(x+y\right)^2}\)
MB
1
23 tháng 1
\(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{2\cdot\left(x+y\right)}{6\left(x+y\right)^2}\)
\(\dfrac{1}{2x+2y}=\dfrac{1}{2\left(x+y\right)}=\dfrac{3\left(x+y\right)}{6\left(x+y\right)^2}\)
\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{1}{\left(x+y\right)^2}=\dfrac{6}{6\left(x+y\right)^2}\)
KN
29 tháng 11 2019
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
KN
29 tháng 11 2019
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 9 2023
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
12 tháng 10 2021
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
15 tháng 11 2017
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) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)
L
1 tháng 8 2021
a) \(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
1 tháng 8 2021
b) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+y^3\right)+\left(3x^2+3xy^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+3xy-1\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)
\(\dfrac{1}{3x+xy}=\dfrac{1}{x\left(y+3\right)}=\dfrac{\left(x+y\right)^2}{x\left(y+3\right)\left(x+y\right)^2}\)
\(2x+2y=2\left(x+y\right)=\dfrac{2\left(x+y\right)\cdot x\left(y+3\right)\left(x+y\right)^2}{x\left(y+3\right)\left(x+y\right)^2}\)
\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{3x+xy}{x\left(y+3\right)\left(x+y\right)^2}\)