2x+2 .2x=192
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a) 2x+2 - 2x= 192
2x . 22 - 2x . 1 = 192
2x . ( 22 - 1) = 192
2x . 3 = 192
2x = 192 : 3
2x = 64 = 26
=> x = 6
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1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
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\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)
\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)
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( 2x -280 +20 ) : 192 =0
<=>2x-280+20=0
<=>2x-280=-20
<=>2x=260
<=>x=130
( 2x - 280 + 20 ) : 192 = 0
=> 2x - 280 + 20 = 0
=> 2x - 280 = - 20
=> 2x = 260
=> x = 130
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1, \(4x=5y\)
mà \(y-2x=-5\)
\(\Rightarrow x=\frac{y+5}{2}\)
\(\Rightarrow\left(\frac{y+5}{2}\right).4=5y\)
\(\Rightarrow\frac{4y+20}{2}=5y\)
\(\Rightarrow2y+10=5y\)
\(\Rightarrow10=3y\)
\(\Rightarrow y=\frac{10}{3}\)
\(\Rightarrow x=\frac{y+5}{2}=\frac{\frac{10}{3}+5}{2}=\frac{\frac{25}{3}}{2}=\frac{25}{6}\)
Vậy \(x=\frac{25}{6};y=\frac{10}{3}\)
b, \(\frac{x}{3}=\frac{y}{4}\)
mà \(xy=192\)
Gọi \(x=3k\)
\(y=4k\)
\(\Rightarrow3k.4k=192\)
\(\Rightarrow12.k^2=192\)
\(\Rightarrow k^2=\frac{192}{12}\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k^2=4^2\)
\(\Rightarrow k=4\)
\(\Rightarrow x=3k=3.4=12\)
\(\Rightarrow y=4k=4.4=16\)
Vậy \(x=12;y=16\)
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