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1) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
2) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-2+1\)
\(\Leftrightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
3) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+2=x+4\)
\(\Leftrightarrow2=4\)
\(\Rightarrow x\in\varnothing\)
4) \(\left(2x-3\right)^5=-243\)
\(\Leftrightarrow\left(2x-3\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow2x-3=-3\)
\(\Leftrightarrow2x=-3+3\)
\(\Leftrightarrow2x=0\)
\(\Rightarrow x=0\)
\(\left(x-2\right)^2=1\)
\(=>\left(x-2\right).\left(x-2\right)=1\)
\(+TH1\)\(\left(x-2\right)=1\)
\(=>1.1=1\left(TM\right)\)
\(+TH2\)\(\left(x-2\right)=-1\)
\(=>\left(-1\right).\left(-1\right)=1\left(TM\right)\)
\(=>x=1;-1\)
1)(x-3)2=0
=>x-3=0
x=0+3
x=3
2)(2x+1)2=4=22
=>2x+1=2
2x=2-1
2x=1
x=1/2
3)(2x-3)3=8=23
=>2x-3=2
2x=2+3
2x=5
x=5/2
4)(x+1/2)4=1/16=(1/4)4
=>x+1/2=1/4
x=1/4-1/2=1/4-2/4
x=-1/4
5)9,27<=3x<=243
32<=3x<=35 (vì x thuộc Z nên làm tròn số 9,27)
=>x thuộc{3;4}
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
\(\left(x-2\right)^2=1\)
=> \(\left(x-2\right)^2=1^2\)
=> \(\left(x-2\right)^2=\left(-1\right)^2\)
=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1+2\\x=\left(-1\right)+2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{3;1\right\}.\)
\(\left(2x-1\right)^3=-8\)
=> \(\left(2x-1\right)^3=\left(-2\right)^3\)
=> \(2x-1=-2\)
=> \(2x=\left(-2\right)+1\)
=> \(2x=-1\)
=> \(x=\left(-1\right):2\)
=> \(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}.\)
\(\left(2x-3\right)^5=-243\)
=> \(\left(2x-3\right)^5=\left(-3\right)^5\)
=> \(2x-3=-3\)
=> \(2x=\left(-3\right)+3\)
=> \(2x=0\)
=> \(x=0:2\)
=> \(x=0\)
Vậy \(x=0.\)
Chúc bạn học tốt!
a) (x - 2)2 = 1
=> x - 2 = 1 hoặc x - 2 = -1
x = 3 ; x = 1
Vậy x = 3; x = 1
b) (2x - 1)3 = -8
=> 2x - 1 = -2
2x = -1
x = \(\frac{-1}{2}\)
Vậy x = \(\frac{-1}{2}\)
c) (2x - 3)5 = -243
=> (2x - 3)5 = (-3)5
=> 2x - 3 = -3
2x = 0
x = 0
Vậy x = 0