Tìm x:
a) x - (\(\frac{50x}{100}+\frac{25x}{200}\)) = \(11\frac{1}{4}\)
b) (x - 5) . \(\frac{30}{100}\) = \(\frac{200x}{100}\) + 5
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\(c,\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow(x+1)(x+1)=2.8\)
\(\Rightarrow(x+1)^2=16\)
\(\Rightarrow(x+1)^2=4^2\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
\(a,x-(\frac{50x}{100}+\frac{25x}{200})=11\frac{1}{4}\)
\(\Rightarrow x-\frac{50x+25x}{100}=\frac{45}{4}\)
\(\Rightarrow\frac{100x}{100}-\frac{75x}{100}=\frac{45}{4}\)
\(\Rightarrow\frac{100x-75x}{100}=\frac{1125}{100}\)
\(\Rightarrow25x=1125\)
\(\Rightarrow x=45\)
\(x-\left(\frac{50x}{100}+\frac{25x}{100}\right)=\frac{45}{4}\)
\(x-\left(\frac{50x+25x}{100}\right)=\frac{45}{4}\)
\(x-\frac{75x}{100}=\frac{45}{4}\)
\(x-x\times\frac{3}{4}=\frac{45}{4}\)
\(x\times\left(1-\frac{3}{4}\right)=\frac{45}{4}\)
\(x\times\frac{1}{4}=\frac{45}{4}\)
\(x=\frac{45}{4}\div\frac{1}{4}\)
\(x=45\)
\(x-\left(\frac{50x}{100}+\frac{25x}{200}\right)=\frac{45}{4}\)
\(x-\left(\frac{100x}{200}+\frac{25x}{200}\right)=\frac{45}{4}\)
\(x-\left(\frac{100x+25x}{200}\right)=\frac{45}{4}\)
\(x-\frac{125x}{200}=\frac{45}{4}\)
\(x-x\frac{5}{8}=\frac{45}{4}\)
\(x\left(1-\frac{5}{8}\right)=\frac{45}{4}\)
\(x.\frac{3}{8}=\frac{45}{4}\)
\(x\)=\(\frac{45}{4}:\frac{3}{8}\)
\(x\)=\(30\)
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