Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow2^3< 2^x< 2^4\)
=>3<x<4
mà x là số nguyên
nên \(x\in\varnothing\)
b: \(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
=>12-x=4
hay x=8
c: \(\Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{2}{5}\right)^2=\left(\dfrac{2}{5}\right)^5\)
=>x>5
d: \(\Leftrightarrow3x-1=-4\)
=>3x=-3
hay x=-1
a) \(8< 2^x\le2^9.2^{-5}\)
\(\Leftrightarrow2^3< x\le2^{9-5}\)
\(\Leftrightarrow2^3< 2^x\le2^4\)
\(\Leftrightarrow3< x\le4\Leftrightarrow x=4\)
b) \(27< 81^3:3^x< 243\)
\(\Leftrightarrow3^2< \left(3^4\right)^3:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12}:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12-x}< 3^5\)
\(\Leftrightarrow2< 12-x< 5\)
\(\Leftrightarrow\hept{\begin{cases}x=8\\x=9\end{cases}}\)
(3.x)^2 : 3^3=243
9.x^2 : 27=243
9.x^2=243.27
9.x^2=6561
x^2=6561:9
x^2=729
X=27
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a,\(8< 2^x\le2^9.2^{-5}\)
\(2^3< 2^x\le2^4\)
\(\Rightarrow x=4\)
b, \(27< 81^3.3^x< 243\)
\(3^3< 3^{12-x}< 3^5\)
\(\Rightarrow3< 12-x< 5\)
12-x=4
x=8
c,\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^2\)
\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^5\)
\(\Rightarrow x>5\)
x=6;7;8........
a: \(\Leftrightarrow2^3< 2^x< 2^4\)
=>3<x<4
mà x là số nguyên
nên \(x\in\varnothing\)
b: \(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
=>12-x=4
hay x=8
c: \(\Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{2}{5}\right)^2=\left(\dfrac{2}{5}\right)^5\)
=>x>5
d: \(\Leftrightarrow3x-1=-4\)
=>3x=-3
hay x=-1