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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) \(8< 2^x\le2^9.2^{-5}\)
\(\Leftrightarrow2^3< x\le2^{9-5}\)
\(\Leftrightarrow2^3< 2^x\le2^4\)
\(\Leftrightarrow3< x\le4\Leftrightarrow x=4\)
b) \(27< 81^3:3^x< 243\)
\(\Leftrightarrow3^2< \left(3^4\right)^3:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12}:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12-x}< 3^5\)
\(\Leftrightarrow2< 12-x< 5\)
\(\Leftrightarrow\hept{\begin{cases}x=8\\x=9\end{cases}}\)
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: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
Ta có :
\(8< 2^x\le2^9.2^{-5}\)(1)
Xét :
\(2^9.2^{-5}=2^9.\frac{1}{2^5}=2^4\)(2)
Thay (2) vào (1) ta có :
\(\Rightarrow2^3< 2^x\le2^4\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
\(a)8< 2^x\le2^9.2^{-5}\)
\(\Leftrightarrow2^3< 2^x\le2^9.\dfrac{1}{2^5}\)
\(\Leftrightarrow2^3< 2^x\le2^4\)
\(\Leftrightarrow x=4\)
\(b)27< 81^3:3^x< 243\)
\(\Leftrightarrow3^3< \left(3^4\right)^3:3^x< 3^5\)
\(\Leftrightarrow3^3< 3^{12}:3^x< 3^5\)
\(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
\(\Leftrightarrow12-x=4\)
\(\Leftrightarrow x=8\)
\(P/s:\)\(Bạn\) \(tự\) \(kết\) \(luận\) \(nha\)