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Lethuha
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Zi Heo
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Lê Hải Yến
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Duong hanh trang
8 tháng 11 2018 lúc 10:07

(x-1)(x-2)(x+2)-(x-3)\(^3\)

=(x-1)(x\(^2\)-4)-(x-3)\(^3\)

(xy-1)(xy-2)-(xy-2)\(^2\)

=(xy-2)(xy-1-xy+2)

=xy-2

Tuyết Ly
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ILoveMath
5 tháng 1 2022 lúc 22:26

\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)

\(b,\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x+3}{x^2-1}\\ =\dfrac{3x^2+4x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+4x+1-x^2+2x-1}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{x^2+2x-3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{2x^2+6x-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x^2+3x\right)+\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)

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

Trần Thị Thanh Bình
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Nguyễn Tuấn
23 tháng 5 2016 lúc 20:44

a)2/x+4x+2/x^2-1

b)

trần thị minh thu
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Zi Heo
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Nguyễn Hoàng Minh
9 tháng 11 2021 lúc 16:39

\(=\dfrac{\left(x-1\right)^3}{xy\left(x-1\right)-\left(x-1\right)}=\dfrac{\left(x-1\right)^3}{\left(xy-1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{xy-1}\left(xy\ne1;x\ne1\right)\)

Nguyễn Thiện Nhân
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trần thảo lê
28 tháng 12 2017 lúc 20:26

\(M=\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

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

\(M=\dfrac{1}{x-1}-\dfrac{1}{x-5}\)

\(M=\dfrac{x-5-x+1}{\left(x-5\right)\left(x-1\right)}=-\dfrac{4}{x^2-6x+5}\)

Duong Thi Nhuong
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Đặng Minh Triều
18 tháng 6 2016 lúc 9:53

ĐK: x khác 1 ; -1

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

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

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

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