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(2x+1)4=(2x+1)6
=>(2x+1)6-(2x+1)4=0
=>(2x+1)4.[(2x+1)2-1]=0
=>(2x+1)4=0 =>x=-1/2
hoặc (2x+1)2-1=0
=>(2x+1)2=1
=>2x+1=1
=>x=0
vậy x=-1/2;0
Th1: 2x+1= -1
2x=-2
x=-1
Th2: 2x+1=0
2x=-1
x=-1/2
Th3: 2x+1=1
2x=0
x=0
(2x+1)4 = (2x+1)6
=>1=(2x+1)6 :(2x+1)4 =
=>1=(2x+1)2
=>2x+1=1 hoặc -1
=>2x+1=1=>x=0;2x-1=-1=>x=-1
Bài 1: (1/2x - 5)20 + (y2 - 1/4)10 < 0 (1)
Ta có: (1/2x - 5)20 \(\ge\)0 \(\forall\)x
(y2 - 1/4)10 \(\ge\)0 \(\forall\)y
=> (1/2x - 5)20 + (y2 - 1/4)10 \(\ge\)0 \(\forall\)x;y
Theo (1) => ko có giá trị x;y t/m
Bài 2. (x - 7)x + 1 - (x - 7)x + 11 = 0
=> (x - 7)x + 1.[1 - (x - 7)10] = 0
=> \(\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1\end{cases}}\)
=> x = 7
hoặc : \(\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)
=> x = 7
hoặc : \(\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
Bài 3a) Ta có: (2x + 1/3)4 \(\ge\)0 \(\forall\)x
=> (2x +1/3)4 - 1 \(\ge\)-1 \(\forall\)x
=> A \(\ge\)-1 \(\forall\)x
Dấu "=" xảy ra <=> 2x + 1/3 = 0 <=> 2x = -1/3 <=> x = -1/6
Vậy Min A = -1 tại x = -1/6
b) Ta có: -(4/9x - 2/5)6 \(\le\)0 \(\forall\)x
=> -(4/9x - 2/15)6 + 3 \(\le\)3 \(\forall\)x
=> B \(\le\)3 \(\forall\)x
Dấu "=" xảy ra <=> 4/9x - 2/15 = 0 <=> 4/9x = 2/15 <=> x = 3/10
vậy Max B = 3 tại x = 3/10
(2x-1)6=(2x-1)8
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=1\\2x-1=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\2x=2\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=1\end{array}\right.\)
+2x -1 = 1 => 2x =2 => x =1
+ 2x -1 = -1 => 2x=0 => x =0
Vậy x = 0 ; 1
(2x-1)^6=(2x-1)^8'
=> (2x-1)=1 hoặc (2x-1)=0
(2x-1)=1 =>2x=2 =>x=1
(2x-1)=0 =>2x=1 =>x=0,5
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 !!!
x< 0=> -x+(-x)-1= 3
-2x= 4
x= -2
0<x<1=> x+(-x)-1=3
0-1=3
-1=3( vô lí )
x>1=> x+x+1=3
2x+1= 3
2x= 2
x= 1
vậy x= -2;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
Vì 2 số có cơ số bằng nhau và số mũ khác nhau nhưng giá trị bằng nhau nên cơ số chỉ có thể là 0 hoặc 1
Nếu giá trị = 0 => x = (0 - 1) : 2 = -0,5
Nếu giá trị = 1 => x = (1 - 1) : 2 = 0
Kết luận : x = -0,5 hoặc = 0