\(2^x.2^{x+1}=128\)
a; 2^x . 16 = 128
b; 3^x : 9 = 27
c; ( 2 . x + 1) ^3 =27
d; ( x . 2 )^2= ( x . 2)^4
e; 2^x . 4 = 128
f; 85^x . 17^x= 625
g; ( x - 2)^4 = 4096
h; 3^x-1+5 . 3^x-1 = 162
a,2^x*16=128
2^x=8
2^x=2^3
suy ra x=3
b,3^x/9=27
3^x/3^2=3^3
3^x=3^5
suy ra x=5
c,(2*x+1)^3=27
(2*x+1)^3=3^3
suy ra 2*x+1=3
suy ra 2x=2
suy ra x=1
d,(x*2)^2=(x*2)^4
suy ra (x*2)^4-(x*2)^2=0
(x*2)^2*((x*2)^2-1)=0
suy ra (x*2)^2=0 hoac (x*2)^2-1=0
+(x*2)^2=0 +(x*2)^2-1=0
suy ra x*2=0 (x*2)^2=1
suy ra x=0 suy ra x*2=1 hoac -1
suy ra x=1/2 hoac -1/2
e,2^x*4=128
2^x=32
2^x=2^5
suy ra x=5
f,85^x*17^x=625
1445^x=625(vo ngiem)
g,(x-2)^4=4096
(x-2)^4=8^4
suy ra x-2=8
suy ra x=10
h,3^(x-1)+5*3*(x-1)=162
6*3^(x-1)=162
3^(x-1)=27
3^(x-1)=3^3
suy ra x-1=3
suy ra x=4
\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+2}=\dfrac{1}{128}\)
\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+2}=\dfrac{1}{128}\\ \left(\dfrac{1}{2}\right)^x.\left[1+\left(\dfrac{1}{2}\right)^2\right]=\dfrac{1}{128}\\ \left(\dfrac{1}{2}\right)^x.\dfrac{5}{4}=\dfrac{1}{128}\\ \left(\dfrac{1}{2}\right)^x=\dfrac{1}{128}:\dfrac{5}{4}=\dfrac{1}{128}.\dfrac{4}{5}=\dfrac{4}{640}=\dfrac{1}{160}\)
Thầy thấy số lẻ quá....
`#3107.101107`
\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+2}=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\left[1+\left(\dfrac{1}{2}\right)^2\right]=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\left(1+\dfrac{1}{4}\right)=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\dfrac{5}{4}=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\dfrac{1}{128}\div\dfrac{5}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\dfrac{1}{160}\)
Bạn xem lại đề.
\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+2}=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^x\cdot\left(\dfrac{1}{2}\right)^2=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\left[1+\left(\dfrac{1}{2}\right)^2\right]=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\left(1+\dfrac{1}{4}\right)=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x\cdot\dfrac{5}{4}=\dfrac{1}{128}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\dfrac{1}{128}:\dfrac{5}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\dfrac{1}{160}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^x=\dfrac{1}{2^5\cdot5}\)
Bạn xem lại đề nhé!
1. 2^x.4=128
2 x^15=x
3. 16^x<128
4. 5^x.5^x+1.5^x+2< hoặc = 100...0 (18 chữ số 0) chia 2^18
5. 2^x.(2^2)^2=(2^3)^2
6. (x^5)^10=x
X*[1/2+1/4+1/8+1/16+1/32+1/64+1/128]=127/128
X x (1/2+1/4+1/8+1/16+1/32+1/64+1/128) = 127/128
X x 127/128 = 127/128
X = 127/128 : 127/128
X = 1
Tìm số tự nhiên x biết
a) 2^x .4 = 128 b) x^15 = x c) 16^x < 128 d) 5^x . 5^ x+1 . 5^ x+2 < hoặc =1 000...0(18c/s0) :2^18
e) 2^x. (2^2)^2= (2^3)^2 f) (x^5)^10 = x
Tìmx,y biết x^2-25=0 b, (x+1)^2=4 c, (x-3)^3-64=0 d, 2^x+1-128=0 e, x^2 +(y-2)^2=0
a: =>(x-5)(x+5)=0
=>x=5 hoặc x=-5
b: \(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow x\in\left\{1;-3\right\}\)
c: =>x-3=4
hay x=7
d: =>x+1=7
hay x=6
e: =>x=0 và y-2=0
hay x=0 và y=2
Bài 1 : Tìm tất cả các số nguyên x , y thỏa mãn :
x2y - x + xy = 6
Bài 2 : Tính nhanh nếu có thể :
d) 512 . ( 2 - 128 ) - 128 . ( - 512 )
d, 512. ( 2 - 128 ) - 128 . ( - 512 )
= 512 . ( - 126 ) - ( - 128 ) . 512
= 512. ( - 126 + 128 )
= 512. 2
= 1024
512 . ( 2 - 128 ) - 128 . ( -512 )
= 512 . -126 - 128 . 512
= 512 . ( -126 + 128 )
= 512 . 2
= 1024
2^x = 128
3^x+1 = 81
( x - 2 )^3 = 1000
2x=128
=> x=7
3x+1=81
=>x+4=4
=>x=3
(x-2)3=1000
=>x-2=10
=>x=12
câu 1 : 2x = 128
=> x = 7
vậy x = 7
Câu 2 : 3x + 1 = 81
=> 3x = 81 - 1
=> 3x=80
Mà 33 = 27 và 34 = 81
mà 3 và 4 là 2 số tự nhiên liên tiếp nên : 3x = 80 ( vô lý )
Vậy x không có gía trị
câu 3
vì 103=1000
=> x - 2 = 10
=> x = 10 +2
=> x = 12
vậy x = 12
\(2^x=128\Rightarrow2^x=2^7\Rightarrow x=7\)
\(3^{x+1}=81\Rightarrow3^{x+1}=3^4\Rightarrow x+1=4\Rightarrow x=4-1=3\)
\(\left(x-2\right)^3=10^3\Rightarrow x-2=10\Rightarrow x=10+2=8\)
a, tim x; 2*(x-1)6=128
b,x2=x
=> (x - 1)2 = 64 => (x - 1)2 = 82 => x - 1 = 8 => x = 9
b/ => x2 - x = 0
=> x(x - 1) = 0
=> x = 0 hoặc x - 1 = 0 => x = 1
a) 2 . (x - 1)2 = 128
(x - 1)2 = 64 = + 8
=> x = 9 hoặc x = -7
b) x2 = x
<=> x \(\in\) {0; -1; 1}
- 3 ( - x + 1 ) - 2 ( x + 7 ) - 32 = - ( - 2x - 9 ) - 128