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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: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a)\(5^x.\left(5^3\right)^2=625\)
\(5^x.5^6=5^4\)
\(5^x=5^{-2}\)
\(x=-2\)
b)\(27< 81^3:3^x< 243\)
\(3^3< \left(3^4\right)^3:3^x< 3^5\)
\(3^3< 3^{12}:3^x< 3^5\)
\(3^{12}:3^x=3^4\)
\(3^x=3^3\)
\(x=3\)
c)\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(5x+1=\frac{6}{7}\)
\(5x=\frac{-1}{7}\)
\(x=\frac{-1}{35}\)
d)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)
\(x-\frac{2}{9}=\frac{4}{9}\)
\(x=\frac{6}{9}=\frac{2}{3}\)
\(5^x.\left(5^3\right)^2=625\)
\(\Rightarrow5^x.5^6=5^4\)
\(\Rightarrow5^{x+6}=5^4\Rightarrow x+6=4\Rightarrow x=-2\)
Đề sai rồi bạn : Phải là :
\(5^x:\left(5^3\right)^2=625\)
\(\Rightarrow5^x:5^6=5^4\)
\(\Rightarrow5^{x-6}=5^4\)
\(\Rightarrow x-6=4\Rightarrow x=10\)
Nhứng nếu đề đúng thì bạn có thể lấy KQ trên
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) \(x=\frac{2}{3}\)
1) \(\left|x-\frac{3}{5}\right|< \frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}< -\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}+\frac{3}{5}\\x< \frac{-1}{3}+\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x< \frac{5}{15}+\frac{9}{15}\\x< \frac{-5}{15}+\frac{9}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
vay \(\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
2) \(\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)
\(\left|x+\frac{11}{2}\right|>5,5\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>\frac{11}{2}\\x+\frac{11}{2}>-\frac{11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{11}{2}-\frac{11}{2}\\x>\frac{-11}{2}-\frac{11}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
vay \(\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
3) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\left|x-\frac{7}{5}\right|>\frac{2}{5}\) va \(\left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{5}>\frac{2}{5}\\x-\frac{7}{5}>\frac{-2}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{2}{5}+\frac{7}{5}\\x>\frac{-2}{5}+\frac{7}{5}\end{cases}}\)va \(\orbr{\begin{cases}x-\frac{7}{5}< \frac{3}{5}\\x-\frac{7}{5}< \frac{-3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{3}{5}+\frac{7}{5}\\x< \frac{-3}{5}+\frac{7}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>\frac{9}{5}\\x>1\end{cases}}\)va \(\orbr{\begin{cases}x< 2\\x< \frac{4}{5}\end{cases}}\)
vay ....