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a) \(2^{x+2}-2^2=96\)
<=> \(2^x.2^2-2^x=96\)
<=> \(2^x\left(4-1\right)=96\)
<=> \(3.2^x=96\)
<=> \(2^x=32\)
<=> \(2^x=2^5\)
<=> x = 5
b, \(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=11\frac{1}{4}\)
\(\Rightarrow x-\left(\frac{1x}{2}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\left(\frac{4x}{8}+\frac{1x}{8}\right)=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{8x}{8}-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\Rightarrow x=\frac{45}{4}\div\frac{3}{8}=30\)
Vậy x = 30
\(\Leftrightarrow\left(x-5\right)\cdot\dfrac{3}{10}=\dfrac{x}{5}+5\)
=>3/10x-3/2=1/5x+5
=>1/10x=5+3/2=10/2+3/2=13/2
=>x=13/2:1/10=13/2x10=65
1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
=>3/10(x-5)=2x+5
=>3/10x-3/2=2x+5
=>-17/10x=5+3/2=6/2+3/2=11/2
=>x=-55/17