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18 tháng 12 2021
a: \(=\dfrac{4x-2+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2x-1}{2x\left(2x-1\right)}\)
20 tháng 12 2021
c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)
22 tháng 7 2023
`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`
`= (a^2+9)/(a^2-9)`
`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`
`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`
`= (2x+2)/(x(x-1)(x+1)`
`= 2/(x(x-1))`
7 tháng 1 2022
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
AD
23 tháng 7 2023
\(a,=\dfrac{5x}{4y^3}\times\left(\dfrac{-20y}{x^4}\right)=\dfrac{-100xy}{4x^4y^3}=\dfrac{-25}{x^3y^2}\\ b,=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+4\right)}\times\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)
\(c,=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\times\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x+3\right)^2.\left(x^2+2x+4\right)}\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 7 2023
a) \(\dfrac{5x}{4y^3}:\left(-\dfrac{x^4}{20y}\right)=\dfrac{5x}{4y^3}\cdot\left(-\dfrac{20y}{x^4}\right)=\dfrac{5\cdot-5}{y^2\cdot x^3}=\dfrac{-25}{x^3y^2}\)
b) \(\dfrac{x^2-16}{x+4}:\dfrac{2x-8}{x}=\left(x-4\right)\cdot\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)
c) \(\dfrac{2x+6}{x^3-8}:\dfrac{\left(x+3\right)^3}{2x-4}=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x^2+2x+4\right)\left(x+3\right)^2}\)
KC
30 tháng 12 2017
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
H
22 tháng 7 2023
\(a,\dfrac{x}{x+3}+\dfrac{2-x}{x+3}\\ =\dfrac{x+2-x}{x+3}\\ =\dfrac{2}{x+3}\\b,\dfrac{x^2y}{x-y}-\dfrac{xy^2}{x-y}\\ =\dfrac{x^2y-xy^2}{x-y}\\ =\dfrac{xy\left(x-y\right)}{x-y}\\ =xy\\ c,\dfrac{2x}{2x-y}+\dfrac{y}{y-2x}\\=\dfrac{2x}{2x-y}-\dfrac{y}{2x-y}\\ =\dfrac{2x-y}{2x-y}\\ =1 \)
22 tháng 7 2023
`a, x/(x+3) + (2-x)/(x+3) = (x+2-x)/(x+3) = 2/(x+3)`
`b, (x^2y)/(x-y) - (xy^2)/(x-y) = (x^2y-xy^2)/(x-y) = (xy(x-y))/(x-y)= xy`
`c, (2x)/(2x-y) - (y)/(2x-y)`
`= (2x-y)/(2x-y) = 1`
\(=\dfrac{2-2x+6}{x-4}=\dfrac{-2x+8}{x-4}=-2\)