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=(3^2)^3.(2^3)^7/(2.3)^9.(2^4)^2
=3^6.2^21/2^9.3^9.2^8
=1.2^4/1.3^3.1
=16/27
\(\left(2x+3\right)^2+\left(3x-2\right)^4=0\)
vì \(\left(2x+3\right)^2\ge0;\left(3x-2\right)^4\ge0\)
nên\(\Rightarrow\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x+3=0\\3x-2=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}\)
Theo đề bài ta có :
\(7^{2x}+7^{2x+2}=2450\)
\(\Rightarrow7^{2x}.\left(1+7^2\right)=2450\)
\(\Rightarrow7^{2x}.50=2450\)
\(\Rightarrow7^{2x}=2450:50\)
\(\Rightarrow7^{2x}=49\)
\(\Rightarrow7^{2x}=7^2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=2:2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
Ta có: \(\frac{9^3.8^7}{6^9.16^2}=\frac{\left(3^2\right)^3.\left(2^3\right)^7}{2^9.3^9.\left(2^4\right)^2}\)=\(\frac{3^6.2^{21}}{2^9.3^9.2^8}=\frac{3^6.2^{21}}{2^{17}.3^9}=\frac{16}{27}\)
\(\frac{^{9^3}.^{^{8^7}}}{^{ }6^9.^{ }16^9}=\frac{729.2097152}{10077696.256}=\frac{1528823808}{2579890176}=\frac{16}{27}\)
\(7^{2x}+7^{2x+2}=2450\)
\(7^{2x}+7^{2x}.7^2=2450\)
\(7^{2x}+7^{2x}.49=2450\)
\(7^{2x}\left(1+49\right)=2450\)
\(7^{2x}.50=2450\)
\(7^{2x}=2450:50\)
\(7^{2x}=49\)
\(7^{2x}=7^2\)
\(2x=2\)
=> \(x=1\)
Vậy \(x=1\)
72x+72x+2=2450
=>72x(1+72)=2450
=>72x.50=2450
=>72x=2450/50
=>72x =49=72
=>2x=2
=>x=1
Vậy x=1
Theo đề bài :
72x + 72x+2 = 2450
=> 72x . (1 + 72) = 2450
=> 72x . (1 + 49) = 2450
=> 72x . 50 = 2450
=> 72x = 49
=> 2x = 2
=> x = 1
Cho mình !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
72x+72x+2=2450
72x x (1+72)=2450
72x x 50 =2450
72x=2450:50
72x=49=72
suy ra 2x=2
suy ra x=1