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Đức Anh Ramsay
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Nguyễn Lê Phước Thịnh
21 tháng 2 2021 lúc 22:08

a) Ta có: \(\dfrac{1-x}{x^2-2x+1}+\dfrac{x+1}{x-1}\)

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

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

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

b) Ta có: \(\dfrac{2x}{3y^4z}\cdot\left(-\dfrac{4y^2z}{5x}\right)\cdot\left(-\dfrac{15y^3}{8xz}\right)\)

\(=\dfrac{2x\cdot4y^2z\cdot15y^3}{3y^4z\cdot5x\cdot8xz}\)

\(=\dfrac{120xy^5z}{120x^2y^4z^2}=\dfrac{y}{xz}\)

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M A S T E R🍎『LⓊƒƒỾ 』⁀...
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Nguyễn Hoàng Minh
19 tháng 10 2021 lúc 19:38

\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)

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Vũ Nguyên Trần Thế
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nthv_.
14 tháng 9 2021 lúc 9:12

\(a.\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=-\dfrac{25}{6}\)

\(b.-\dfrac{1}{12}-\dfrac{5}{4}+\dfrac{1}{3}=-\dfrac{1}{12}-\dfrac{15}{12}+\dfrac{4}{12}=-1\)

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_Jun(준)_
14 tháng 9 2021 lúc 9:14

a) \(\dfrac{-7}{2}+\dfrac{3}{4}-\dfrac{17}{12}=\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=\dfrac{-42+9-17}{12}=-\dfrac{50}{12}=-\dfrac{25}{6}\)

b)\(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)=-\dfrac{2}{24}-\left(\dfrac{63}{24}-\dfrac{8}{24}\right)\)

\(=-\dfrac{2}{24}-\dfrac{63}{24}+\dfrac{8}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)

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Đức Anh Ramsay
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Nguyễn Phương Linh
17 tháng 2 2021 lúc 13:05

ĐKXĐ: \(a\ne1\)

a. \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)

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

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

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

\(=-\dfrac{1}{a^2+a+1}\)

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Nguyễn Lê Phước Thịnh
17 tháng 2 2021 lúc 13:07

a) Ta có: \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)

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

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

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

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

b) Ta có: \(x-\dfrac{xy}{x+y}-\dfrac{x^3}{x^2y^2}\)

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

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

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

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

 

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

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

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

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

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

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Lê Phương Linh
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Nguyễn Xuân Thành
26 tháng 8 2023 lúc 7:41

a) \(\dfrac{-4}{3}-\dfrac{2}{7}+\dfrac{4}{3}\)

\(=\left(\dfrac{-4}{3}+\dfrac{4}{3}\right)-\dfrac{2}{7}\)

\(=0-\dfrac{2}{7}=\dfrac{-2}{7}\)

b)  \(\dfrac{1}{2}+\dfrac{-4}{5}-\dfrac{3}{10}\)

\(=\dfrac{5}{10}+\dfrac{-8}{10}-\dfrac{3}{10}\)

\(=\dfrac{5+\left(-8\right)-3}{10}=\dfrac{-3}{5}\)

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Kiều Vũ Linh
26 tháng 8 2023 lúc 7:45

b) 1/2 + (-4/5) - 3/10

= 5/10 - 8/10 - 3/10

= -6/10

= -3/5

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DSQUARED2 K9A2
26 tháng 8 2023 lúc 8:24

a, -4/3 - 2/7 + 4/3

= ( -4/3 + 4/3 ) - 2/7

= 0 - 2/7

= -2/7

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Vũ Thảo Anh
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Lê Mai Tuyết Hoa
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Nguyễn Hoàng Minh
15 tháng 12 2021 lúc 18:44

\(a,=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{x-y}{xy\left(y-x\right)}=\dfrac{-1}{xy}\\ b,=\dfrac{x+3-x-4}{x-2}=\dfrac{-1}{x-2}\)

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Bành Thị Kem Trộn
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Vô danh
21 tháng 3 2022 lúc 9:20

\(a,\dfrac{-8}{15}+\dfrac{13}{30}-\dfrac{5}{12}=\dfrac{-32}{60}+\dfrac{26}{60}-\dfrac{25}{60}=-\dfrac{31}{60}\\ b,\dfrac{3}{2}.\dfrac{7}{2}+\left(\dfrac{-5}{6}+\dfrac{1}{10}:\dfrac{11}{30}\right)=\dfrac{21}{4}+\left(\dfrac{-5}{6}+\dfrac{3}{11}\right)=\dfrac{21}{4}+\dfrac{-37}{66}=\dfrac{619}{132}\)

\(c,\dfrac{-20}{21}.\dfrac{22}{35}+\dfrac{-20}{21}.\dfrac{13}{35}+\dfrac{-22}{21}=\dfrac{-20}{21}\left(\dfrac{22}{35}+\dfrac{13}{35}\right)+\dfrac{-22}{21}=\dfrac{-20}{21}.1+\dfrac{-22}{21}=\dfrac{-20}{21}+\dfrac{-22}{21}=\dfrac{-42}{21}=-2\)

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