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a ) \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left(2x-1\right)^4=3^4\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow x=2\)
Vậy \(x=2.\)
b ) \(\left(x-1\right)^5=-32\)
\(\Leftrightarrow\) \(\left(x-1\right)^5=-2^5\)
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-1\)
Vậy \(x=-1.\)
c ) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(1-2x+1\right)\left(1+2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2-2x\right).2x\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\2-2x=0\\2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x=2\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)
Vậy ...............
a) (2x-1)4=81
\(\Leftrightarrow\)\(\left[\begin{array}{} (2x-1)^4=(3)^4\\ (2x-1)^4=(-3)^4 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} 2x-1=3\\ 2x-1=-3 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} 2x=3+1\\ 2x=-3+1 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} 2x=4\\ 2x=-2 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} x=4:2\\ x=-2:2 \end{array}\right.\)
\(\Rightarrow\)\(\left[\begin{array}{} x=2\\ x=-1 \end{array}\right.\)
Vậy x=2 hoặc x=-1
b) (x-1)5= -32
\(\Leftrightarrow\)\( (x-1)^5=(-2)^5 \)
\(\Rightarrow\)\( (x-1)=-2 \)
\(\Rightarrow\)\( x=-2+1 \)
\(\Rightarrow\)\( x=-1 \)
Vậy x=-1
c) ( 2x-1)6= ( 2x-1)8
\(\Leftrightarrow\) (2x-1)6=(2x-1)8.
\(\Leftrightarrow\)(2x-1)8-(2x-1)6=0.
\(\Leftrightarrow\)(2x-1)6)[(2x-1)2-1]=0.
\(\Leftrightarrow\)(2x-1)6(2x-1+1)(2x+1+1)=0.
\(\Leftrightarrow\)(2x-1)62x(2x+2)=0.
\(\Leftrightarrow\)(2x-1)6<=>2x(2x-1)=0.\(\Rightarrow x=\dfrac{1}{2}\)
hoặc 2x=0\(\Rightarrow\)x=0
hoặc 2x+2=0\(\Rightarrow\)2x=-2\(\Leftrightarrow\)x=-2:2\(\Leftrightarrow\)x=-1
Vậy x=\(\dfrac{1}{2}\)hoặc x=0 hoặc x=-1
Chúc bạn học tốt !!!
(2x-1)6 = (2x-1)8
=> 2x-1 thuộc {0; 1; -1}
TH1: 2x-1 = 0
=> x = 1/2
TH2: 2x-1 = 1
=> x = 1
TH3: 2x-1 = -1
=> X = 0
KL:............................
(2x-1)^6=(2x-1)^8
==>((2x-1)^8)/(2x-1)^6=1
(2x-1)^2=1
4x^2-4x+1=1
4x(x-1)=0
==> 2 nghiem x=0 va x=1
đúng thì cho ****
(2x-1)6=(2x-1)8
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=1\\2x-1=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\2x=2\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=1\end{array}\right.\)
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
=>\(\left(2x-1\right)^6:\left(2x-1\right)^6=\left(2x-1\right)^8:\left(2x-1\right)^6\)
=>\(1=\left(2x-1\right)^2\)
=>\(\left(-1\right)^2=\left(2x-1\right)^2\) hoặc \(1^2=\left(2x-1\right)^2\)
=>2x-1=-1 hoặc 2x-1=1
+)Nếu 2x-1=-1
=>2x=0
=>x=0
+)Nếu 2x-1=1
=>2x=2
=>x=1
Vậy x=0 hoặc x=1
a/ \(\left(2x-4\right)^4=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-4\right)^4=3^4\\\left(2x-4\right)^4=\left(-3\right)^4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=3\\2x-4=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy .......
b/ \(\left(x-1\right)^5=-32\)
\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-1\)
Vậy ............
a) \(\left(2x-4\right)^4=81\\ \left(2x-4\right)^4=3^4\\\Rightarrow2x-4=3\\ 2x=3+4\\ 2x=7\\ x=7:2\\ x=\dfrac{7}{2} \)
Vậy \(x=\dfrac{7}{2}\)
b) \(\left(x-1\right)^5=-32\\ \left(x-1\right)^5=\left(-2\right)^5\\ \Rightarrow x-1=-2\\ x=-2+1\\ x=-1\)
Vậy \(x=-1\)
c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \)
Suy ra không tìm được x
+2x -1 = 1 => 2x =2 => x =1
+ 2x -1 = -1 => 2x=0 => x =0
Vậy x = 0 ; 1
c) 5x+5x+2=650
=> 5x+5x52 = 650
=> 5x ( 1+ 25 ) = 650
=> 5x = 650 / 26
=> 5x = 25 = 52
=> x = 2
\(\left(2x-1\right)^6=\left(1-2x\right)^8\)
<=>\(\left(1-2x\right)^6=\left(1-2x\right)^8\)
<=>\(\left(1-2x\right)^8-\left(1-2x\right)^6=0\)
<=>\(\left(1-2x\right)^6\left(\left(1-2x\right)^2-1\right)=0\)
<=>1-2x=0 hoặc (1-2x)^2-1=0
<=>x=1/2 hoặc x=1
a, (2x-1)^6= (2x-1)^8
=>2x-1=1 hoặc 2x-1=0
2x=2 hoặc 2x=1
x=1 hoặc x=1/2
b, 5^x + 5^x+2= 650
5x+5x.52=650
5x(1+52)=650
5x.26=650
5x=650:26
5x=25
5x=52
=>x=2
b, 5^x + 5^x+2 = 650
5^x ( 1 + 5^2) = 650
5^x . 26 = 650
5^x =650:26
5^x = 25
5^x = 5^2
x = 2
(2x-1)8-(2x-1)6=0
(2x-1)6((2x-1)2-1)=0
=>(2x-1)6=0 hoặc ((2x-1)2-1)=0
xét từng trường hợp =0 nhế
(2x-1)6 = (2x-1)8
=> (2x-1)6 - (2x-1)8 = 0
(2x-1)6.[ 1 - (2x-1)2 ] = 0
=> (2x-1)6 = 0 => 2x - 1= 0 => 2x = 1 => x = 1/2
1 - (2x-1)2 = 0 => (2x-1)2 = 1 = 12 = (-1)2 => 2x - 1 =1 => 2x = 2 => x = 1
2x - 1 = - 1 => 2x = 0 =>x = 0
KL:...