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Nguyễn Lương Bảo Tiên
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Nguyễn Minh Quang 123
27 tháng 11 2015 lúc 22:24

\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)

 

 

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kagamine rin len
27 tháng 11 2015 lúc 22:25

1/ (x+1)(x+2) +1/ (x+2)(x+3) +1/ (x+3)(x+4) +1/ (x+4)(x+5)

=1/x+1 -1/x+2 +1/x+2 -1/x+3 +1/x+3 -1/x+4 +1/x+4 -1/x+5

=1/x+1 -1/x+5

=4/(x+1)(x+5)

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Nguyễn Thị Thùy Dương
27 tháng 11 2015 lúc 22:27

 

\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)

\(=\frac{1}{x+1}-\frac{1}{x+5}=\frac{x+5-x-1}{\left(x+1\right)\left(x+5\right)}=\frac{4}{\left(x+1\right)\left(x+5\right)}\)

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êfe
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Nguyễn Thị Mỹ Lệ
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Kẻ Vô Danh
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Hà Hoài Thư
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Ryan Nguyễn
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lucky tomato
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Nguyễn Việt Hoàng
31 tháng 3 2020 lúc 9:06

\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{\left(x+3\right)\left(x+2-2x-2\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{\left(x+3\right)\left(-x\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{-x^2-3x+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

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Trịnh Xuân Ngọc
1 tháng 4 2020 lúc 9:18

ĐKXD: x\(\ne\)-1,-2,-3

Ta có

\(\frac{1}{\left(x+1\right)\left(x+2\right)}\)-\(\frac{2}{\left(x+2\right)^2}\)+\(\frac{1}{\left(x+2\right)\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(x+3+x+1\right)-2\left(x^2+4x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{\left(x+2\right)\left(2x+4\right)-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{2x^2+8x+8-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

=\(\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

Chúc bạn học tốt

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Nobi Nobita
3 tháng 4 2020 lúc 15:52

\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne-2\\x\ne-3\end{cases}}\)

\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}+\frac{-2\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+2\right)^2}+\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+3\right)\left(x+2\right)^2}\)

\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{x^2+5x+6-2\left(x^2+4x+3\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)\(=\frac{x^2+5x+6-2x^2-8x-6+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

\(=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)

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Hoàng Quang Kỳ
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Nguyễn Văn Tuấn
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nguyen van chien
17 tháng 1 2019 lúc 0:16

tra loi nhanh di ae

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