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a,\(x+5=20-\left(12-7\right)\)
\(x+5=15\)
\(x=10\)
b,\(15+\left(x-4\right)=20\)
\(x-4=5\)
\(x=9\)
a) \(60-3.\left(x-2\right)=51\)
\(\Rightarrow3.\left(x-2\right)=60-51=9\)
\(\Rightarrow x-2=9:3=3\)
\(\Rightarrow x=3+2=5\)
b) \(\left(105-x\right):2^5=3^{0+1}=3^1=3\)
\(\Rightarrow\left(105-x\right):32=3\)
\(\Rightarrow105-x=3.32=96\)
\(\Rightarrow x=105-96=9\)
c) \(x+5=20-\left(12-7\right)\)
\(\Rightarrow x+5=20-5\)
\(\Rightarrow x+5=15\)
\(\Rightarrow x=15-5=10\)
a) \(2x+5=20-\left(12-7\right)\)
\(2x+5=20-5\)
\(2x+5=15\)
\(2x=15-5\)
\(2x=10\)
\(x=10:2\)
\(x=5\)
Vậy \(x=5\)
b) \(\left|x\right|=\left|-5\right|\)
\(\Rightarrow\left|x\right|=5\)
\(\Rightarrow x=\pm5\)
Vậy \(x\in\left\{5;-5\right\}\)
3:
\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)
4:
=>2/3:x=-2-1/3=-7/3
=>x=-2/3:7/3=-2/7
5:
AC=CB=12/2=6cm
IB=6/2=3cm
\(\frac{3}{4}x-\frac{2}{5}x=-\frac{7}{20}\)
\(\left(\frac{3}{4}-\frac{2}{5}\right)x=-\frac{7}{20}\)
\(\frac{7}{20}x=-\frac{7}{20}\)
\(x=-\frac{7}{20}:\frac{7}{20}\)
\(x=-1\)