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2^x*2^2-2^X=96
2^x(4-1)=96
2^x*3=96
2^x=96/3
2^x=32
2^x=2^5
Vậy x=5
\(2^{X+2}-2^X=96\)
\(=>2^X.\left(2^2-1\right)=96\)
\(=>2^X.3=96\)
\(=>2^X=96:3\)
\(=>2^X=32\)
\(=>2^X=2^5\)
\(=>X=5\)
2x+2 - 2x =96
2x * 22 -2x *1=96
2x *(22 -1)=96
2x *(4-1)=96
2x *3=96
2x =96:3
2x =32
2x =25
x =5
Vậy x=5
2x+2 - 2x =96
2x.22-2x=96
2x.(22-1)=96
2x.3=96
2x=96:3
2x=32
2x=25
=>x=5
2x+2-2x=96
=>2x(22-1)=96
=>2x.3=96
=>2x=96:3
=>2x=32=25
=>x=5
vậy x=5
=> 2^x ( 2^2 + - 1 ) = 96
=> 3.2^x = 96
=> 2^x = 96 : 3
=> 2^x = 32
=> 2^ x = 2^5
=> x = 5
2^x + 2^x+1=96
2^x+2^x.2=96
2^x(1+2)=96
2^x.3=96
2^x=96:3
2^x=32
2^x=2^5
=>x=5
vậy x=5
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a.
\(7^{x+2}+2\times7^{x-1}=345\)
\(7^x\times7^2+2\times7^x\div7=345\)
\(7^x\times\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\times\frac{345}{7}=345\)
\(7^x=345\div\frac{345}{7}\)
\(7^x=345\times\frac{7}{345}\)
\(7^x=7\)
\(x=1\)
b.
\(2^{x+2}-2^x=96\)
\(2^x\times\left(2^2-1\right)=96\)
\(2^x\times3=96\)
\(2^x=\frac{96}{3}\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
\(a,7^{x+2}+2.7^{x-1}=345\Rightarrow7^x.49+\frac{2}{7}.7^x=345\Rightarrow7^x\left(49+\frac{2}{7}\right)=345\Rightarrow7^x.\frac{345}{7}=345\Rightarrow7^x=345:\frac{345}{7}=7^1\Rightarrow x=1\)
\(b,2^{x+2}-2^x=96\Rightarrow2^x.4-2^x=96\Rightarrow2^x\left(4-1\right)=96\Rightarrow2^x.3=96\Rightarrow2^x=96:3=32\Rightarrow2^x=2^5\Rightarrow x=5\)
\(a,7^{x+2}+2.7^{x-1}=345=>7^{x-1+3}+2.7^{x-1}=345=>7^{x-1}.7^3+2.7^{x-1}=345\)
\(=>\left(7^3+2\right).7^{x-1}=345=>345.7^{x-1}=345=>7^{x-1}=1=7^0=>x-1=0=>x=1\)
\(b,2^{x+2}-2^x=96=>2^x.2^2-2^x=96=>2^x.\left(4-1\right)=96=>2^x.3=96=>2^x=32=2^5=>x=5\)
a)
(2x+1)2=25
=> \(\left[\begin{array}{nghiempt}2x+1=5\\2x+1=-5\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}2x=4\\2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
d)
(x-1)3=-125
=> x-1=-5
=> x=-4
còn câu b và c bạn viết đề rõ hơn nha
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
\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x.4-2^x=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x=32\)
\(\Rightarrow x=5\)
Vậy x = 5
\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x.4-2^x=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x.3=96\)
\(\Rightarrow2^x=32\)
\(\Rightarrow x=5\)