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1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
![](https://rs.olm.vn/images/avt/0.png?1311)
5^x + 5^ ( x + 2 ) = 650
5x + 5x . 52 = 650
5x .( 1 + 25 ) = 650
5x . 26 = 650
5x = 650 : 26
5x = 25
5x = 52
=> x = 2
Vậy x = 2
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\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)
Tìm x biết:
a,x-5/7=1/9
b,2x/5=6/2x+1
c,11/8+13/6=85/x
d,2x-2/11=1.1/5
e,x/15=3/5+-2/3
f,x/182=-6/14.35/91
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{52}{63}\)
b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)
2\(x\).(2\(x\) + 1) = 30
4\(x^2\)+ 2\(x\) - 30 = 0
4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0
(4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0
4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0
2 (\(x\) + 3).(2\(x\) - 5) = 0
\(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}
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1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)
=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)
=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)
c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)
=>2x=4
=>x=4/2=2
d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)
=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)
=>x=-2
e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)
=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)
=>\(\dfrac{7}{3}x=2,4\)
=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)
f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)
=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)
=>\(x=-\dfrac{2}{15}\cdot45=-6\)
g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)
=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)
=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)
=>x=10/2=5
h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)
=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)
=>4x=7
=>\(x=\dfrac{7}{4}\)
a) x= 2, 3...8
a: Ta có: \(36\le6^x\le1296\)
\(\Leftrightarrow2\le x\le4\)
hay \(x\in\left\{2;3;4\right\}\)
b: Ta có: \(100< 5^{2x-1}< 5^6\)
\(\Leftrightarrow2x-1\in\left\{3;5\right\}\)
\(\Leftrightarrow2x\in\left\{4;6\right\}\)
hay \(x\in\left\{2;3\right\}\)