v.v

\(5^5:5^3=5^{5-3}=5^2=25\)

Công thức:

\(a^m:a^n=a^{m-n}\) 

-1

\(3.\left(\dfrac{1}{2}\right)^0+\left[2.5-\dfrac{1}{2}\right]^2-2^3\)

\(=3.\left(\dfrac{1}{2}\right)^0+\left[\dfrac{5}{2}-\dfrac{1}{2}\right]^2-2^3\)

\(=3.\left(\dfrac{1}{2}\right)^0+\left[2\right]^2-2^3\)

\(=3.1+4-8\)

\(=3+4-8\)

\(=7-8\)

\(=7+\left(-8\right)\)

\(=1\)

\(\left(6\times5+7-37\right)\times\left(1+2+3+4+...+9+10\right)\)

\(=\left(30+7-37\right)\times\left(1+2+3+...+9+10\right)\)

\(=\left(37-37\right)\times\left(1+2+3+...+10\right)\)

\(=0\times\left(1+2+3+...+10\right)\)

\(=0\)

Vì đó là quy định chung!

\(\left(20:x\right)-18:3=4\)

\(\left(20:x\right)-6=4\)

\(20:x=4+6\)

\(20:x=10\)

      \(x=20:10\)

      \(x=2\)

Vậy...

\(3.\left(x-7\right)+48=12.5\)

\(3.\left(x-7\right)+48=60\)

\(3.\left(x-7\right)=60-48\)

\(3.\left(x-7\right)=12\)

     \(x-7=12:3\)

     \(x-7=4\)

     \(x=4+7\)

     \(x=11\)

Vậy...

\(\dfrac{15}{19}.\dfrac{38}{5}< x< \dfrac{67}{15}+\dfrac{56}{15}\)

Ta có:

\(\dfrac{15}{19}.\dfrac{38}{5}\)

\(=\dfrac{3}{1}.\dfrac{2}{1}\)

\(=6\)

\(\dfrac{67}{15}+\dfrac{56}{15}=\dfrac{123}{15}=\dfrac{41}{5}=8,2\)

\(=>x\in\left\{7\right\}\)

Vậy...

Cậu ghi TK hộ mình đc k?

\(37.23+63.12+37.15+26.63\)

\(=37.\left(23+15\right)+63.\left(12+26\right)\)

\(=37.38+63.38\)

\(=\left(37+63\right).38\)

\(=100.38\)

\(3,54.73+0,23.25+3,54.27+0,17.25\)

\(=3,54.\left(73+27\right)+25.\left(0,23+0,17\right)\)

\(=3,54.100+25.0,4\)

\(=354+10\)

\(=364\)

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