(2x-√9/4)^2=1/√625
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\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\sqrt{\dfrac{1}{25}}\\2x-\dfrac{3}{2}=-\sqrt{\dfrac{1}{25}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{10}\\2x=\dfrac{13}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)
Vậy ..
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\\ \Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(\dfrac{4}{9}\right)^x=\left(\dfrac{3}{2}\right)^{-5}\\ \Leftrightarrow\left(\dfrac{2}{3}\right)^{2x}=\left(\dfrac{2}{3}\right)^5\\ \Rightarrow x=\dfrac{5}{2}\)
Vậy....
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a: =>2x+5=4
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)
=>(3x-4)(3x-5)(3x-3)=0
hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)
c: \(\Leftrightarrow3^{x+1}=3^{2x}\)
=>2x=x+1
=>x=1
d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)
=>2x+3=2x-10
=>0x=-13(vô lý)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)
b: =>|17x-5|=|17x+5|
=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5
=>34x=0
hay x=0
c: =>|3x+4|=|4x-18|
=>4x-18=3x+4 hoặc 4x-18=-3x-4
=>x=22 hoặc 7x=14
=>x=22 hoặc x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
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Bài làm
a) x² - 3 = 22
=> x² = 25
=> x = + 5
Vậy x = + 5
b) 2x³ + 5 = -11
2x³ = -16
x³ = -8
x = -2
Vậy x = -2
c) ( x + 2 )² = 81
=> x + 2 = 9
=> x = 7
Vậy x = 7
d) ( 2x + 1 )² = 25
=> 2x + 1 = 5
=> 2x = 4
=> x = 2
Vậy x = 2
e) 5x + 2 = 625
5x = 623 ( vô lí )
g) ( 2x - 3 )² = 36.
=> 2x - 3 = 6
=> 2x = 9
=> x = 4,5
Vậy x = 4,5
h) ( 2x - 1 )³ = -8
=> 2x - 1 = -2
=> 2x = -1
=> x = -1/2
Vậy x = -1/2
i) ( x - 1 )x + 2 = ( x - 1 )x + 6
=> [ (x - 1 )x - ( x - 1 )x ] = 6 - 2
=> 0 = 4 ( vô lí )
Vậy x thuộc rỗng.
k) x² + x = 0
=> x( x + 1 ) = 0
=> x = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
Vậy x = 0 hoặc x = -1
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a) |x-1|-2x=5
\(\Leftrightarrow\) x - 1 - 2x = 5 (đk: x\(\ge\)1 ) (1)
hoặc -x + 1 - 2x = 5 ( đk: x < 1) (2)
(1): \(\Leftrightarrow\) x - 2x = 5 + 1
\(\Leftrightarrow\) - x = 6
\(\Leftrightarrow\) x = -6 (ko thỏa mãn)
(2): \(\Leftrightarrow\) -x + 1 - 2x = 5
\(\Leftrightarrow\) -3x = 4
\(\Leftrightarrow\) x = \(\frac{-4}{3}\) (thỏa mãn)
Vậy: x = \(\frac{-4}{3}\)
b) |9-7x|=5x-3
\(\Leftrightarrow\) 9 - 7x = 5x - 3 (đk: \(\le\) \(\frac{9}{7}\) ) (1)
hoặc -9 + 7x = 5x - 3 (đk: x > \(\frac{9}{7}\) ) (2)
(1): \(\Leftrightarrow\) -7x-5x = -3-9 \(\Leftrightarrow\) -12x = -12 \(\Leftrightarrow\) x = 1 ( thỏa mãn)
(2): \(\Leftrightarrow\) 7x - 5x = -3 + 9 \(\Leftrightarrow\) 2x = 6 \(\Leftrightarrow\) x = 3 ( thỏa mãn)
Vậy: x= 1; 3
c) 5x+2 = 625
\(\Leftrightarrow\) 5x . 52 = 54
\(\Leftrightarrow\) 5x = \(\frac{5^4}{5^2}\)= 52
\(\Leftrightarrow\) x = 2
d) (2x - 3)2 = 36
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-3=6\left(đk:x\ge\frac{3}{2}\right)\text{}\Leftrightarrow x=4,5\left(TMĐK\right)\\2x-3=-6\left(đk:x< \frac{3}{2}\right)\Leftrightarrow x=-1,5\left(TMĐK\right)\end{matrix}\right.\)
Vậy: x = 4,5; -1,5
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)