Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2^{x+2}-96=2^x\)\(\Leftrightarrow2^{x+2}-2^x=96\)
\(\Leftrightarrow2^x\left(2^2-1\right)=96\)\(\Leftrightarrow2^x.3=96\)
\(\Leftrightarrow2^x=32=2^5\)\(\Leftrightarrow x=5\)
Vậy \(x=5\)
\(2^{x+2}-2^x=96\)
\(\Leftrightarrow2^x\cdot2^2-2^x=96\)
\(\Leftrightarrow2^x\left(2^2-1\right)=96\)
\(\Leftrightarrow2^x\cdot3=96\)
\(\Leftrightarrow2^x=96\div3\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy x = 5
2^(x+2) - 2^x = 96
<=> (2^x)(2^2) -2^x = 96
(2^x)[(2^2) -1)] = 96
2^x = 96/3 = 32
2^5 = 32
Đáp số:
x = 5
chúc bn hok tốt
Đặt \(\dfrac{x}{3}=\dfrac{y}{2}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=2k\end{matrix}\right.\)
Ta có: \(x^2y^2=96\)
\(\Leftrightarrow6k^2=96\)
\(\Leftrightarrow k^2=16\)
Trường hợp 1: k=4
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=12\\y=2k=8\end{matrix}\right.\)
Trường hợp 2: k=-4
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-12\\y=2k=-8\end{matrix}\right.\)
\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x.4-2^x=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x=32\)
\(\Rightarrow x=5\)
Vậy x = 5
\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x.4-2^x=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x.3=96\)
\(\Rightarrow2^x=32\)
\(\Rightarrow x=5\)
a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow x+100=0\)
hay x=-100
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 ....