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a) Ta có: \(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)
\(\Leftrightarrow x^2-10x+18-x^2=12\)
\(\Leftrightarrow-10x+18=12\)
\(\Leftrightarrow-10x=-6\)
hay \(x=\frac{3}{5}\)
Vậy: \(x=\frac{3}{5}\)
b) Ta có: \(7x\left(x-2\right)-5\left(x-1\right)=7x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5-7x^2-3=0\)
\(\Leftrightarrow-19x+2=0\)
\(\Leftrightarrow-19x=-2\)
hay \(x=\frac{2}{19}\)
Vậy: \(x=\frac{2}{19}\)
tìm x, y biết \(\frac{\left(2x+1\right)}{5}\)=\(\frac{\left(4y-5\right)}{9}\)=\(\frac{2x+4y-4}{7x}\)
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a) (5x + 1)2 = 36/49
(5x + 1)2 = (6/7)2
suy ra: 5x + 1= 6/7 hoặc 5x +1 = -6/7
5x = 6/7 - 1 5x = -6/7 -1
5x = -1/7 5x = -13/7
x = -1/7 : 5 x = -13/7 : 5
x = -1/35 x = -13/35
a, \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
=> 5x + 1 = \(\frac{6}{7}\)
\(\Rightarrow5x=\frac{6}{7}-1\)
\(\Rightarrow5x=\frac{-1}{7}\)
\(\Rightarrow x=\frac{-1}{7}:5=\frac{-1}{35}\)
b,, \(\left(2x-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\left(2x-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)
\(\Rightarrow2x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}+\frac{1}{2}=1\)
\(\Rightarrow x=1:2=\frac{1}{2}\)
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1
2(\(\frac{3}{4}\)-5x)=\(\frac{4}{5}\)-3x
=> \(\frac{6}{4}-10x=\frac{4}{5}-3x\)
=>\(-10x+3x=\frac{4}{5}-\frac{6}{4}\)
=> \(x=\frac{1}{10}\)
2 .
\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
=>\(\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
=>\(11x=\frac{1}{6}\)
=>x=\(\frac{1}{66}\)
3.
\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
=>\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)
=>\(-2x=\frac{-2}{3}\)
=>\(\frac{1}{3}\)
4. câu 4 ko hiểu bạn ơi
![](https://rs.olm.vn/images/avt/0.png?1311)
khỏi chép lại đề ha
- 2 - 4x - 5x + \(\frac{3}{2}\)= \(\frac{7}{4}\)
\(\frac{7}{2}\)- 9x = \(\frac{7}{4}\)
-9x = \(\frac{7}{2}-\frac{7}{4}\)
-9x = \(\frac{7}{4}\)
x = \(\frac{7}{4}:\left(-9\right)\)
x = \(\frac{-7}{36}\)
- 3 - 2x - \(\frac{1}{3}=7x-\frac{1}{4}\)
-2x - 7x = \(\frac{-1}{4}-3+\frac{1}{3}\)
-9x = \(\frac{-35}{12}\)
x = \(\frac{-35}{12}:\left(-9\right)\)
x = \(\frac{35}{108}\)
- \(\frac{-15}{2}\)+ \(\frac{1}{4}\)+ 4x -2 = 1
4x = 1 + \(\frac{15}{2}-\frac{1}{4}+2\)
4x = \(\frac{41}{4}\)
x = \(\frac{41}{4}:4\)
x = \(\frac{41}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)\(-\frac{3}{2}\left(\frac{4}{5}-\frac{2}{3}\right)+x=4\left(x-\frac{2}{3}\right)\)
\(\Leftrightarrow-\frac{3}{2}\cdot\frac{2}{15}+x=4x-\frac{8}{3}\)
\(\Leftrightarrow-\frac{1}{5}+x=4x-\frac{8}{3}\)
\(\Leftrightarrow x-4x=-\frac{8}{3}+\frac{1}{5}\)
\(\Leftrightarrow-3x=-\frac{37}{15}\)
\(\Leftrightarrow x=\frac{37}{45}\)
2)\(2\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow3-2x-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow-2x-7x=-\frac{1}{4}-3+\frac{1}{3}\)
\(\Leftrightarrow-9x=-\frac{35}{12}\)
\(\Leftrightarrow x=\frac{35}{108}\)
3)\(\frac{1}{5}\left(-\frac{3}{5}10\right)+5x=x-\frac{2}{3}\)
\(\Leftrightarrow-\frac{6}{5}+5x=x-\frac{2}{3}\)
\(\Leftrightarrow5x-x=-\frac{2}{3}+\frac{6}{5}\)
\(\Leftrightarrow4x=\frac{8}{15}\)
\(\Leftrightarrow x=\frac{2}{15}\)
4)\(-\frac{3}{2}\left(5-\frac{1}{6}\right)+4\left(x-\frac{1}{2}\right)=1\)
\(\Leftrightarrow-\frac{15}{2}+\frac{1}{4}+4x-2=1\)
\(\Leftrightarrow4x=1+\frac{15}{2}-\frac{1}{4}+2\)
\(\Leftrightarrow4x=\frac{41}{4}\)
\(\Leftrightarrow x=\frac{41}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(7x=9y\) và \(10x-8y=68\)
Có: \(7x=9y\Leftrightarrow\frac{x}{9}=\frac{y}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{9}=\frac{y}{7}=\frac{10x-8y}{90-56}=\frac{68}{34}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.9\\y=2.7\end{cases}}\Rightarrow\hept{\begin{cases}x=18\\y=14\end{cases}}\)
b) Ta có: 7x = 9y => x = 9/7y
Lại có: 10x - 8y = 68
=> 10.9/7.y - 8y = 68
=> 90/7.y - 56/7.y = 68
=> 34/7.y = 68
=> y = 68 : 34/7 = 14
=> x = 9/7.14 = 18
c) Vì (x - 1/2)50 > hoặc = 0; (y + 1/3)40 > hoặc = 0
Mà (x - 1/2)50 + (y + 1/3)40 = 0
=> (x - 1/2)50 = 0; (y + 1/3)40 = 0
=> x - 1/2 = 0; y + 1/3 = 0
=> x = 1/2; y = -1/3
x=6,142857143