\(\dfrac{b}{a-b}\)+\(\dfrac{a}{b-a}\)
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H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(\dfrac{a-1}{a+1}+\dfrac{3-a}{a+1}\)
\(=\dfrac{a-1+3-a}{a+1}\)
\(=\dfrac{2}{a+1}\)
b) \(\dfrac{b}{a-b}+\dfrac{a}{b-a}\)
\(=\dfrac{b}{a-b}+\dfrac{-a}{a-b}\)
\(=\dfrac{b-a}{a-b}\)
\(=-1\)
c) \(\dfrac{\left(a+b\right)^2}{ab}-\dfrac{\left(a-b\right)^2}{ab}\)
\(=\dfrac{\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)+\left(a-b\right)\right]}{ab}\)
\(=\dfrac{4ab}{ab}\)
\(=4\)
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))`
1 tháng 11 2017
a) \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Rightarrow1+\dfrac{b}{a}=1+\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
\(\RightarrowĐPCM\)
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`
19 tháng 11 2021
ĐKXĐ: \(b,d\ne0,c\ne\pm d\)
Áp dụng t/c dtsbn:
\(\dfrac{a^{2k}+b^{2k}}{c^{2k}+d^{2k}}=\dfrac{a^{2k}-b^{2k}}{c^{2k}-d^{2k}}=\dfrac{a^{2k}+b^{2k}+a^{2k}-b^{2k}}{c^{2k}+d^{2k}+c^{2k}-d^{2k}}=\dfrac{2a^{2k}}{2c^{2k}}=\dfrac{a^{2k}}{c^{2k}}\left(1\right)\)
\(\dfrac{a^{2k}+b^{2k}}{c^{2k}+d^{2k}}=\dfrac{a^{2k}-b^{2k}}{c^{2k}-d^{2k}}=\dfrac{a^{2k}+b^{2k}-a^{2k}+b^{2k}}{c^{2k}+d^{2k}-c^{2k}+d^{2k}}=\dfrac{2b^{2k}}{2d^{2k}}=\dfrac{b^{2k}}{d^{2k}}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^{2k}}{c^{2k}}=\dfrac{b^{2k}}{d^{2k}}\Rightarrow\dfrac{a^{2k}}{b^{2k}}=\dfrac{c^{2k}}{d^{2k}}\Rightarrow\dfrac{a}{b}=\pm\dfrac{c}{d}\left(đpcm\right)\)
22 tháng 7 2023
a: \(=\dfrac{3b+4a}{6ab}\)
b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)
c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)
d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
BD
2
TH
6 tháng 1 2021
a)\(=\dfrac{x-2+x+5}{x+3}=\dfrac{2x+3}{x+3}\)
b)ĐKXĐ: \(x+3\ne0\Leftrightarrow x\ne-3\)
6 tháng 1 2021
a) Ta có: \(\dfrac{x-2}{x+3}+\dfrac{x+5}{x+3}\)
\(=\dfrac{x-2+\left(x+5\right)}{x+3}\)
\(=\dfrac{x-2+x+5}{x+3}\)
\(=\dfrac{2x+3}{x+3}\)
b) ĐKXĐ: \(x+3\ne0\)
hay \(x\ne-3\)
20 tháng 8 2023
a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\dfrac{-x^2-4x-4-2x^2+4x-4x^2}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}\)
\(=\dfrac{-7x^2-4}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}=\dfrac{7x^2+4}{\left(x+2\right)\left(x-3\right)}\)
b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)
8 tháng 11 2017
\(\text{a) }\dfrac{x^2+2}{x^3-1}+\dfrac{x}{x^2+x+1}+\dfrac{1}{1-x}\\ =\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x}{x^2+x+1}-\dfrac{1}{x-1}\\ =\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}-\dfrac{\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{x^2+2+x\left(x-1\right)-\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{x^2-2x+1}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x-1\right)^2}{\left(x^2+x+1\right)\left(x-1\right)}\\ =\dfrac{x-1}{x^2+x+1}\\ \)
\(\text{b) }\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\\ =\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{x\left(y-x\right)}\\ =\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\\ =\dfrac{x^2}{y\left(x-y\right)x}-\dfrac{\left(2x-y\right)y}{x\left(x-y\right)y}\\ =\dfrac{x^2-\left(2x-y\right)y}{xy\left(x-y\right)}\\ =\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\\ =\dfrac{\left(x-y\right)^2}{xy\left(x-y\right)}\\ =\dfrac{x-y}{xy}\)
`#3107.101107`
`b/(a - b) + a/(b - a)`
`= b/(a - b) - a/(a - b)`
`= (b - a)/(a - b)`
`= -(a - b)/(a - b)`
`= -1`