\(\left(\frac{1}{1}-X\right)^2-1\frac{3}{9}=1\frac{4}{9}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
=>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
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 :>
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x = \frac{{ - 9}}{{10}}\).
Bài của mình bị thiếu 1 trường hợp nên bạn tham khảo bài của bạn kia nhé. :<<
\(\left(1-x\right)^2-1\frac{3}{9}=1\frac{4}{9}\)
=> \(\left(1-x\right)^2=1\frac{4}{9}+1\frac{3}{9}=\frac{25}{9}\)
=> \(\left(1-x\right)^2=\left(\frac{5}{3}\right)^2\)
=> \(\orbr{\begin{cases}1-x=\frac{5}{3}\\1-x=-\frac{5}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=\frac{8}{3}\end{cases}}\)