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\(a)\)\(\left(\frac{3}{5}\right)^{2x+1}=\frac{81}{625}\)
\(\Leftrightarrow\)\(\left(\frac{3}{5}\right)^{2x+1}=\left(\frac{3}{5}\right)^4\)
\(\Leftrightarrow\)\(2x+1=4\)
\(\Leftrightarrow\)\(x=\frac{3}{2}\)
Vậy \(x=\frac{3}{2}\)
\(b)\)\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)^3=\frac{32}{243}\)
\(\Leftrightarrow\)\(\left(\frac{2}{3}\right)^{x+3}=\left(\frac{2}{3}\right)^5\)
\(\Leftrightarrow\)\(x+3=5\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(c)\)\(\left(2x-1\right)^2=\left(2x-1\right)^3\)
\(\Leftrightarrow\)\(\left(2x-1\right)^3-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-1-1\right)=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(2x-1\right)^2=0\\2x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=1\)
Chúc bạn học tốt ~
a, \(A=\frac{15\left|x+1\right|+32}{6\left|x+1\right|+8}\Rightarrow2A=\frac{30\left|x+1\right|+64}{6\left|x+1\right|+8}=5+\frac{24}{6\left|x+1\right|+8}=5+\frac{12}{3\left|x+1\right|+4}\)
Ta thấy \(\left|x+1\right|\ge0\) với mọi x
\(\Rightarrow3\left|x+1\right|\ge0\Rightarrow3\left|x+1\right|+4\ge4\Rightarrow0< \frac{12}{3\left|x+1\right|+4}\le3\)
\(\Rightarrow5< A\le8\)
Suy ra GTLN của A là 8 khi |x+1|=0 hay x=-1
VẬY GTLN của A là 8 khi x=-1
câu a thui còn câu b mk chưa có bít làm
bn k cho mk nha
Bạn Aquarius bài sai rùi
Bạn ấy ghi 2A=... mà chưa =>A=...
sao bạn kết luận hay thế?
\(\left(2x-4\right)^4=81\)
\(\left(2x-4\right)^4=3^4\)
\(\Rightarrow2x-4=3\)
\(\Rightarrow2x=7\)
\(\Rightarrow x=\frac{7}{2}\)
vay \(x=\frac{7}{2}\)
\(\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(\Rightarrow x-1=-2\)
\(\Rightarrow x=-1\)
vay \(x=-1\)
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)
\(\left(2x-1\right)^6\left(1-2x+1\right)\left(1+2x-1\right)=0\)
\(\left(2x-1\right)^6\left(-2x+2\right)\left(2x\right)=0\)
\(\Rightarrow\left(2x-1\right)^6=0\)hoac \(\Rightarrow\orbr{\begin{cases}-2x+2=0\\2x=0\end{cases}}\)
\(\Rightarrow2x-1=0\) hoac \(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
\(\Rightarrow x=\frac{1}{2}\)hoac \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
5 ( x - 11 ) = 2 ( 2x - 3 )
5x - 55 = 4x - 6
5x - 4x = -6 + 55
x = 49
a/ \(\left(\frac{1}{5}\right)^x=\left(\frac{1}{5^3}\right)^3=\left(\frac{1}{5}\right)^9\Rightarrow x=9\)
b/ \(\left(\frac{3}{5}\right)^x=\left(\frac{3^2}{5^2}\right)^3=\left(\frac{3}{5}\right)^6\Rightarrow x=6\)
c\(2^{3-2x}=\left(2^3\right)^3=2^9\Rightarrow3-2x=9\Rightarrow x=-3\)
d/ \(2^{3x+1}=32^2=\left(2^5\right)^2=2^{10}\Rightarrow3x+1=10\Rightarrow x=3\)
e/ \(3^{6-3x}=81^3=\left(3^4\right)^3=3^{12}\Rightarrow6-3x=12\Rightarrow x=-2\)
\(\left(\frac{1}{5}\right)^x=\left(\frac{1}{125}\right)^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left[\left(\frac{1}{5}\right)^3\right]^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left(\frac{1}{5}\right)^9\Leftrightarrow x=9\)
\(\left(\frac{3}{5}\right)^x=\left(\frac{9}{25}\right)^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left[\left(\frac{3}{5}\right)^2\right]^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^6\Leftrightarrow x=6\)
\(2^{3-2x}=8^3\Leftrightarrow2^{3-2x}=\left(2^3\right)^3\Leftrightarrow2^{3-2x}=2^9\Leftrightarrow3-2x=9\)
\(\Leftrightarrow2x=3-9\Leftrightarrow2x=-6\Leftrightarrow x=\left(-6\right):2\Leftrightarrow x=-3\)
Các phép còn lại làm tương tự bn nha !