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20 tháng 8 2020

Bài của mình bị thiếu 1 trường hợp nên bạn tham khảo bài của bạn kia nhé. :<<

20 tháng 8 2020

\(\left(1-x\right)^2-1\frac{3}{9}=1\frac{4}{9}\)

=> \(\left(1-x\right)^2=1\frac{4}{9}+1\frac{3}{9}=\frac{25}{9}\)

=> \(\left(1-x\right)^2=\left(\frac{5}{3}\right)^2\)

=> \(\orbr{\begin{cases}1-x=\frac{5}{3}\\1-x=-\frac{5}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=\frac{8}{3}\end{cases}}\)

21 tháng 6 2017

Ta có : (1/4)2 - 2/3x = (2/3)2

=> 1/16 - 2/3x = 4/9

=> 2/3x = 1/16 - 4/9

=> 2/3x = -55/144

=> x = -55/144 . 3/2

=> x = -55/96

22 tháng 8 2019

=>2/9*(x-9/4)+1/2=41/14+1/3

=>2/9*(x-9/4)=137/42-1/2

=> x-9/4=58/21:2/9

=>x=87/7+9/4

=>x=411/28

  nếu sai thì trách cái máy tính chứ đừng có trách mk

22 tháng 8 2019

T lười chép đề bài lắm :)

\(\frac{2}{9}\cdot\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}\cdot\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}\cdot\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

\(\frac{2}{9}\cdot\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}\cdot\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}\cdot\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

\(x-\frac{9}{4}=\frac{58}{21}:\frac{2}{9}\)

\(x-\frac{9}{4}=\frac{87}{7}\)

\(x=\frac{87}{7}+\frac{9}{4}\)

\(x=\frac{411}{28}\)

Vậy ...

P/s : Chả bt có đúng k :> 

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

giải hệ phương trình 1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\) 2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\) 3 ,...
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giải hệ phương trình

1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)

2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)

3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)

4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)

5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)

6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)

7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)

8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)

9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)

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