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a)2x+2*2x=256
=>22x+2=256
=>22x+2=28
=>2x+2=8
=>2x=6
=>x=3
b)2x+2+2x=80
=>2x(22+1)=80
=>2x*5=80
=>2x=16
=>2x=24
=>x=4
c)2x+2-2x=96
=>2x(22-1)=96
=>2x*3=96
=>2x=32
=>2x=25
=>x=5
a) \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)
⇔ | 3 - 2x | + 3/4 = 11/4
⇔ | 3 - 2x | = 8/4 = 2
⇔ \(\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}}\text{⇔}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)
b) 2x+2 - 2x = 96
⇔ 2x( 22 - 1 ) = 96
⇔ 2x.3 = 96
⇔ 2x = 32
⇔ 2x = 25
⇔ x = 5
c) ( 2x + 5 )3 = -27
⇔ ( 2x + 5 )3 = (-3)3
⇔ 2x + 5 = -3
⇔ 2x = -8
⇔ x = -4
a. \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)
\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\left|-\frac{11}{4}\right|\)
\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\frac{11}{4}\)
\(\Rightarrow\left|3-2x\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)
b. 2x+2 - 2x = 96
<=> 2x . 22 - 2x = 96
<=> 2x ( 22 - 1 ) = 96
<=> 2x . 3 = 96
<=> 2x = 32 = 25
<=> x = 5
c. ( 2x + 5 )3 = - 27
<=> ( 2x + 5 )3 = ( - 3 )3
<=> 2x + 5 = - 3
<=> 2x = - 8
<=> x = - 4
2x + 2 - 2x - 96
=> 2x . 22 - 2x . 1 = 96
=> 2x (22 - 1) = 96
=> 2x . 3= 96
=> 2x = 96 : 3 =32
=> x = 5
2x + 2 - 2x = 96
=> 2x . 22 - 2x . 1 = 96
=> 2x (22 - 1) = 96
=> 2x . 3= 96
=> 2x = 96 : 3 =32
=> x=5.
a) 2x + 2 - 2x = 96
=> 2x . 4 - 2x = 96
=> 2x . (4 - 1) = 96
=> 2x . 3 = 96
=> 2x = 96 : 3
=> 2x = 32
=> 2x = 25
=> x = 5
a) 2x + 2 - 2x = 96
=> 2x. 22 - 2x = 96
=> 2x( 4 - 1 ) = 96
=> 2x = 96 : 3
=> 2x = 25
=> x = 5
a) \(\left(2x+1\right)^2=25\)
=> \(2x+1=5\) và \(2x+1=-5\)
=> \(2x=5-1=4\) và \(2x=-5-1=-6\)
=> \(x=4:2=2\) và \(x=-6:2=-3\)
b) \(\left(x-1\right)^3=-125\)
=> \(x-1=-5\Rightarrow x=-5+1=-4\)
c) \(2^{x+2}-2^x=96\)
=> \(2^x\cdot2^2-2^x\cdot1=96\)
=> \(2^x\left(2^2-1\right)=96\)
=> \(2^x\cdot3=96\Rightarrow2^x=96:2=32\)
=> \(x=5\)
d) \(7^{x+2}+2\cdot7^{x-1}=345\)
=> \(7^x\cdot7^2+2\cdot7^x:7=345\)
=> \(7^x\cdot7^2+2\cdot7^x\cdot\frac{1}{7}=345\)
=> \(7^x\cdot\left(7^2+2\cdot\frac{1}{7}\right)=345\)
=> \(7^x\cdot\frac{345}{7}=345\)
=> \(7^x=345:\frac{345}{7}=7\)
=> \(x=1\)
\(\left(2x+1\right)^2=25\)
\(\left(2x+1\right)^2=5^2=\left(-5\right)^2\)
\(TH1:\left(2x+1\right)^2=5^2\)
\(2x+1=5\)
\(x=\left(5-1\right):2\)
\(x=4\)
\(TH2:\left(2x+1\right)^2=\left(-5\right)^2\)
\(2x+1=-5\)
\(x=\left[\left(-5\right)-1\right]:2\)
\(x=-3\)
Vậy x=2 hoặc x= -3
2x(22-1)=96
2x.4=96
2x=24
=>x ko có giá trị
viết sai đề hả bạn à
\(2^{x+2}-2^x=96\Leftrightarrow2^x.4-2^x.1=96\)
\(\Leftrightarrow2^x\left(4-1\right)=96\Leftrightarrow2^x=32=2^5\Rightarrow x=5\)
Vậy ....