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Những câu hỏi liên quan
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Huỳnh Nguyễn Tuấn Nam
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Nguyễn Thanh Tùng
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Yukino Ayama
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HT.Phong (9A5)
21 tháng 8 2023 lúc 14:18

a) \(\dfrac{2x}{x^2-6x+9}+\dfrac{x-2}{x-3}\) (ĐK: \(x\ne3\))

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

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

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

\(=\dfrac{x^2-3x+6}{x^2-6x+9}\)

b) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)

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

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

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

\(=\dfrac{1}{x^2+x+1}\)

Cỏ dại
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kudo shinichi
13 tháng 12 2018 lúc 22:00

\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)

\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)

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

\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)

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

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

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)

\(=\frac{12}{x}\)

nguyen ngoc son
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bella nguyen
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Nguyễn Lê Phước Thịnh
3 tháng 3 2022 lúc 9:47

a: \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\)

\(=\dfrac{5xy+y^3-x\left(5y-x^2\right)}{x^2y^2}\)

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\)

b: \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)

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

 

Nguyễn Thuỳ Linh
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KAl(SO4)2·12H2O
12 tháng 3 2020 lúc 14:47

\(\frac{6}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\)

\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x}{x-3}+\frac{x}{x+3}\)

\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{6x+5x\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)

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Tran Le Khanh Linh
12 tháng 3 2020 lúc 14:54

\(\frac{6x}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\left(x\ne\pm3\right)\)

\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{6x+5x^2+15x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{6x^2+18x}{\left(x-3\right)\left(x+3\right)}=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)

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Thanhphuoc Nguyen
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Nguyễn Việt Lâm
26 tháng 7 2021 lúc 11:14

\(=\left[\left(-6x\right)+\left(x^2+9\right)\right]\left[\left(-6x\right)-\left(x^2+9\right)\right]\)

\(=\left(-6x\right)^2-\left(x^2+9\right)^2\)

\(=36x^2-\left(x^4+18x^2+81\right)\)

\(=-x^4+18x^2-81\)

\(=-\left(x^4-18x^2+81\right)\)

\(=-\left(x^2-9\right)^2\)

Nguyễn Lê Phước Thịnh
27 tháng 7 2021 lúc 0:42

Ta có: \(\left(x^2-6x+9\right)\left(-x^2-6x-9\right)\)

\(=-\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)

\(=-\left[\left(x-3\right)^2\cdot\left(x+3\right)^2\right]\)

\(=-\left(x^2-9\right)^2\)