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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\hept{\begin{cases}\text{|}0,5x\text{|}=0,5x\\\sqrt{\left(0,5x\right)^2}=0,5x\\\left(0,5x\right)^2=\left(0,5x\right)^2\end{cases}}\)
2, tương tự
\(\hept{\begin{cases}\text{|}-\frac{2}{3}x\text{|}=\frac{2}{3}x\\\sqrt{\left(-\frac{2}{3}x\right)^2}=\frac{2}{3}x\\\left(-\frac{2}{3}x\right)^2=\left(\frac{2}{3}x\right)^2\end{cases}}\)
4, tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
\(2\frac{2}{3}:x=1\frac{7}{9}:2\frac{2}{3}\)
\(\frac{8}{3}:x=\frac{16}{9}:\frac{8}{3}\)
\(\frac{8}{3}:x=\frac{2}{3}\)
\(\frac{8}{3}:\frac{2}{3}=x\)
\(x=4\)
Vậy x = 4
b,
\(-2^3+0,5x=1,5\)
\(-8+0,5x=1,5\)
\(0,5x=1,5+8\)
\(0,5x=9,5\)
\(x=9,5:0,5\)
\(x=19\)
Vậy x = 19
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Rightarrow x^2+3x+x+3=x+4x+0,5x+2\)
\(\Rightarrow x^2+3x+x-x-4x-0,5x=2-3\)
\(\Rightarrow x^2-x=-1\)
\(\Rightarrow x\left(x-1\right)=-1\)
:vvv
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x:
a)-23+0,5x=1,5
-8+0,5x = 1,5
0,5x = 1,5-(-8) = 9,5
x = 9,5:0,5 = 19
b)(−3)x81=−27
(-3)x:81 =-27
(-3)x = -27.81 = -2187
(-3)x = (-3)7
=> x=7
c)112.x−4=0,5
1,5.x = 0,5+4 = 4,5
x = 4,5:1,5 = 3
d)123:x4=6:0,3
\(\frac{5}{3}\):\(\frac{x}{4}\) = 20
\(\frac{x}{4}\) = \(\frac{5}{3}\):20 = \(\frac{1}{12}\)
=> x:4 = \(\frac{1}{12}\)
x = \(\frac{1}{12}\).4 = \(\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
mình viết nhầm, mình sửa lại bài nhé
\(\frac{x+1}{2x+1}-\frac{0,5x+2}{x+3}\)
\(\Rightarrow\) (x+1)(x+3) = (0,5x+2)(2x+1)
\(\Rightarrow\) x2 + 3x + x + 3 = x2 + 0,5x + 4x + 2
\(\Rightarrow\) x2 + 4x + 3 = x2 + 4,5x + 2
\(\Rightarrow\) x2 - x2 + 4x - 4,5x = 2 - 3
\(\Rightarrow\) -0,5x = -1
\(\Rightarrow\) x = \(\frac{-1}{-0,5}\)
\(\Rightarrow\) x = 2
Vậy x = 2
\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Rightarrow\) (x + 1)(x + 3) = (2x + 1)(0,5x + 2)
\(\Rightarrow\) x2 + 3x + x + 3 = x2 + 4x + 0,5x + 2
\(\Rightarrow\) x2 + 3x + x + 3 - x2 - 4x - 0,5x - 2 = 0
\(\Rightarrow\) 0,5x + 1 = 0
\(\Rightarrow\) 0,5x = 0 - 1
\(\Rightarrow\) 0,5x = -1
\(\Rightarrow\) x = -1 : 0,5
\(\Rightarrow\) x = -2
Vậy x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x=4/7 - 1/3=19/21
b) /x-5/=7 -->x-5=7 hoặc x-5=-7
--> x=12 hoặc x= -2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow x.\left(0,5-\frac{2}{3}\right)=\frac{7}{12}\)
\(\Rightarrow x.\left(-\frac{1}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}:\left(-\frac{1}{6}\right)=-\frac{7}{2}\)
\(0,5x-2=x+\frac{2}{3}\)
\(\Leftrightarrow-\frac{1}{2}x=\frac{8}{3}\)
\(\Leftrightarrow x=-\frac{16}{3}\)
#H
0,5x - 2 = x + \(\frac{2}{3}\)
\(-2-\frac{2}{3}=x-0,5x\)
\(-\frac{8}{3}=0,5x\)
\(x=-\frac{16}{3}\)
Thấy đúng k cho tui