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Trần Thị Hảo
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Nguyễn Lê Phước Thịnh
26 tháng 11 2022 lúc 13:59

a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

=0

c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{1}{xyz}\)

 

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Nguyen Thi Ngoc Lan
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Nguyễn Lê Phước Thịnh
28 tháng 11 2022 lúc 13:33

a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)

b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)

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lucky tomato
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Khách vãng lai
31 tháng 3 2020 lúc 0:09

\(\Leftrightarrow\) \(\frac{\left(x-z\right)-\left(x-y\right)}{\left(x-y\right)\left(x-z\right)}\)\(+\frac{\left(y-x\right)-\left(y-z\right)}{\left(y-z\right)\left(y-x\right)}+\frac{\left(z-y\right)-\left(z-x\right)}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)

\(\Leftrightarrow\)\(\frac{1}{x-y}-\frac{1}{x-z}+\frac{1}{y-z}-\frac{1}{y-x}+\frac{1}{z-x}-\frac{1}{z-y}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)

\(\Leftrightarrow\)\(\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}+\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)

tự lm nốt ik

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 Khách vãng lai đã xóa
Sách Giáo Khoa
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Nguyen Thuy Hoa
28 tháng 6 2017 lúc 16:07

Phép cộng các phân thức đại số

Phép cộng các phân thức đại số

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MInemy Nguyễn
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Tuấn
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Nguyễn Thanh Vân
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hattori heiji
21 tháng 11 2017 lúc 22:01

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

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blinkwannable
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trần thảo lê
20 tháng 12 2017 lúc 19:38

a,

\(-\dfrac{x}{\left(x-y\right)\left(z-x\right)}-\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(z-x\right)\left(y-z\right)}\)

\(\dfrac{-x\left(y-z\right)-y\left(z-x\right)-z\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

\(\dfrac{-xy+xz-yz+xy-zx+yz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

= 0

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๖²⁴ʱ乂ų✌й๏✌ρɾ๏༉
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Ngoc An Pham
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Hồng Phúc
6 tháng 2 2021 lúc 9:36

\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\)

\(=\dfrac{z-x+x-y}{\left(x-y\right)\left(z-x\right)}+\dfrac{x-y+y-z}{\left(y-z\right)\left(x-y\right)}+\dfrac{y-z+z-x}{\left(z-x\right)\left(y-z\right)}\)

\(=\dfrac{1}{x-y}+\dfrac{1}{z-x}+\dfrac{1}{y-z}+\dfrac{1}{x-y}+\dfrac{1}{z-x}+\dfrac{1}{y-z}\)

\(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)

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