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nguyễn vũ kim anh
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Kiệt Nguyễn
13 tháng 6 2019 lúc 9:30

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)

\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)

\(\Leftrightarrow\frac{38}{x+3}=16\)

\(\Leftrightarrow x+3=2,375\)

\(\Leftrightarrow x=-0,625\)

 ♡ℓ٥ﻻ ﻉ√٥υ♡
13 tháng 6 2019 lúc 9:38

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)

\(\Leftrightarrow\frac{38}{x+3}=14\)

\(\Leftrightarrow\left(x+3\right)14=38\)

\(\Leftrightarrow14x+42=38\)

\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)

Vậy \(x=-\frac{2}{7}\)

 ♡ℓ٥ﻻ ﻉ√٥υ♡
13 tháng 6 2019 lúc 9:41

TRẻ Trâu 

Trần Hà Nhung
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Phong Linh
1 tháng 8 2018 lúc 7:17

a) Ta có : (x - 5)2 - 16

= (x - 5)2 - 42

= (x - 5 - 4)(x - 5 + 4)

= (x - 1)(x - 9)

b) 25 - (3 - x)2

= 52 - (3 - x)2

= (5 - 3 + x)(5 + 3 - x)

= (x + 2)(8 - x)

Phong Linh
1 tháng 8 2018 lúc 7:21

c) (7x - 4)2 - (2x + 1)2

= (7x - 4 - 2x - 1)(7x - 4 + 2x + 1)

= (5x - 5)(9x - 3)

= 5(x - 1)3(3x - 1)

= 15(x - 1)(3x - 1)

kudo shinichi
1 tháng 8 2018 lúc 9:42

\(49.\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=\left[7.\left(y-4\right)\right]^2-\left[3.\left(y+2\right)\right]^2\)

\(=\left[7.\left(y-4\right)-3.\left(y+2\right)\right].\left[7.\left(y-4\right)+3.\left(y+2\right)\right]\)

\(=\left(7y-28-3y-6\right).\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right).\left(10y-22\right)\)

\(=4.\left(y-17\right).\left(5y-11\right)\)

\(8x^3+\frac{1}{27}=\left(2x\right)^3+\left(\frac{1}{3}\right)^3=\left(2x+\frac{1}{3}\right).\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)\)

\(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right).\left(25+5x^2+x^4\right)=\left(\sqrt{5}-x\right).\left(\sqrt{5}+x\right).\left(25+5x^2+x^4\right)\)

Hien Tran
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Trần Hà Nhung
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Anna
20 tháng 6 2018 lúc 7:57

\(\Leftrightarrow2.\left(\frac{-1}{2}\right).\left(\frac{2}{3}\right)^2-3\left(-\frac{1}{3}\right)^2.\frac{2}{9}:x=3.\left(-\frac{1}{2}\right)-\frac{2}{3}\)

\(\Leftrightarrow-\frac{4}{9}-\frac{1}{3}.\frac{2}{9}:x=-\frac{3}{2}-\frac{2}{3}\)

\(\Leftrightarrow-\frac{4}{6}-\frac{2}{27}:x=-\frac{13}{6}\)

\(\Leftrightarrow\frac{2}{27}:x=-\frac{4}{9}:\frac{-13}{6}\)

\(\Leftrightarrow\frac{2}{27}:x=\frac{31}{18}\)

\(\Leftrightarrow x=\frac{2}{27}:\frac{31}{18}\)

\(\Rightarrow x=\frac{4}{93}\)

Vậy \(x=\frac{4}{93}\)

Thân An Phương
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Member lỗi thời :>>...
14 tháng 7 2021 lúc 15:43

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

Khách vãng lai đã xóa
Trà Huỳnh Anh Khoa
14 tháng 7 2021 lúc 15:44

\(x\in\varnothing\) 

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nguyễn hoàng mai
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Bùi Hồng Duyên
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Edogawa Conan
21 tháng 7 2019 lúc 21:44

a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)

=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

=>  x + 1 = 0

=> x = -1

Edogawa Conan
21 tháng 7 2019 lúc 21:57

b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

=> x - 2021 = 0

=> x = 2021

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

=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)

=> \(-\frac{1}{12}x+6=7\)

=> \(-\frac{1}{12}x=1\)

=> x = -12

nguyễn vũ kim anh
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❤P͟͟.T͟͟↭2K͟͟7➻❥
12 tháng 4 2019 lúc 12:06

cho 3 k 

\(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{10^2}\right)\)

=> \(\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)\)\(...\left(1-\frac{1}{10}\right)\cdot\left(1+\frac{1}{10}\right)\)

=> \(\left(1-\frac{1}{2}\right)\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\cdot\cdot\frac{9}{10}\cdot\frac{10}{11}\)

=> \(\frac{1}{2}\cdot\frac{3\cdot2\cdot4\cdot\cdot\cdot9\cdot10}{2\cdot3\cdot3\cdot\cdot\cdot10\cdot11}=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}\)

Chúc bn học tốt !

❤P͟͟.T͟͟↭2K͟͟7➻❥
12 tháng 4 2019 lúc 12:22

cho mk 3 k nha bn

thanks nhìu

bài này mk ko copy, ko chép mạng, tự nghĩ mất 6 phút . 

có công thức rùi nha !

chúc bn học tốt

Hien Tran
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