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a. x = {3;-3}
b. x thuộc rỗng
c. x2-4=0
x2 = 4
x={2;-2}
d. x2+1=82
x2 =83
x thuộc rỗng
e. (2x)2=6
x thuộc rỗng
f. (x-1)2=9
TH1: x-1=3=>x=4
TH2: x-1=-3=>x=-2
Vậy x={4;-2}
g.(2x+3)2=25
TH1: 2x+3=5=> x=1
Th2: 2x+3=-5=>x=-4
VẬY X={1;-4}
a, x^2= 9
=>\(\sqrt{9}=3\)
b,\(x^2=5=>x=\sqrt{5}\)
c, x^2-4=0
=>x^2=4
=>x=2
d, x^2+1=82
=>x^2=81 =>\(\sqrt{81}=9\)
3, 2x^2=6
=>x= \(\sqrt{6}\)
f, {x-1} ^2=9
=> x-1=3
=>x=2
g{ 2x+3}^2=25
=> 2x+3=5
=>2x=2
=>x=1
a) x^2 = 9 => x=3 hoặc x = -3
b) x^2 = 5 => \(x=\sqrt{5}\)
c) x^2 - 4 = 0
=> x^2 = 4 => x = 2 hoặc x = -2
d) x^2 + 1 = 82
=> x^2 = 81 => x = 9 hoặc x = -9
e) (2x)^2 = 6
=> 4 . x^2 = 6
=> x^2 = 3/2
=> \(x=\sqrt{\frac{3}{2}}\)
f) (x-1)^2 = 9
=> x-1 = 3 hoặc x - 1 = -3
=> x = 4 hoặc -2
g) (2x+3)^2 = 25
=> 2x + 3 = 5 hoặc 2x + 3 = -5
=> x = 1 hoặc x = -4
Ta có:
a, \(x^2=9\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
b, \(x^2=5\Rightarrow\orbr{\begin{cases}x=2,5\\x=-2.5\end{cases}}\)
Các câu còn lại tương tự nhé bn
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
\(a,\dfrac{3}{2}\cdot x-1=\dfrac{1}{2}x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{1}{2}x=-\dfrac{3}{5}+1\)
\(\Rightarrow\left(\dfrac{3}{2}-\dfrac{1}{2}\right)x=-\dfrac{3}{5}+\dfrac{5}{5}\)
\(\Rightarrow x=\dfrac{2}{5}\)
\(b,\dfrac{1}{2}x+\dfrac{1}{2}\left(x-2\right)=\dfrac{3}{4}-2x\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{1}{2}x+2x-1=\dfrac{3}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)
\(\Rightarrow3x=\dfrac{7}{4}\)
\(\Rightarrow x=\dfrac{7}{4}:3\)
\(\Rightarrow x=\dfrac{7}{12}\)
\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
\(d,4^{x-3}+1=17\)
\(\Rightarrow4^{x-3}=17-1\)
\(\Rightarrow4^{x-3}=16\)
\(\Rightarrow4^{x-3}=4^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
#Toru
`3/2 x -1 =1/2x -3/5`
`=> 3/2x -1/2x = -3/5 +1`
`=> 2/2x= -3/5 + 5/5`
`=> x= 2/5`
__
`1/2x +1/2(x-2) = 3/4 -2x`
`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`
`=> 1/2x +1/2x +2x = 3/4 + 1`
`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`
`=> 6/2x = 7/4`
`=> x= 7/4 : 3`
`=>x=7/12`
__
`(x-1/2) -1/4=0`
`=> x-1/2=1/4`
`=> x=1/4 +1/2`
`=> x= 1/4 +2/4`
`=>x=3/4`
__
`4^(x-3) +1=17`
`=> 4^(x-3) =17-1`
`=> 4^(x-3)=16`
`=> 4^(x-3)=4^2`
`=> x-3=2`
`=>x=2+3`
`=>x=5`
\(\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\)
\(\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
=> \(\left(2x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\)
=> \(2x+\dfrac{1}{2}=\pm\dfrac{1}{2}\)
TH1:
\(2x+\dfrac{1}{2}=\dfrac{1}{2}\)
\(2x=\dfrac{1}{2}-\dfrac{1}{2}=0\)
\(x=0\)
TH2:
\(2x+\dfrac{1}{2}=-\dfrac{1}{2}\)
\(2x=-\dfrac{1}{2}-\dfrac{1}{2}\)
\(2x=-1\)
\(x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{0;\dfrac{-1}{2}\right\}\)