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Trả lời :
\(B=\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}=\frac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}=\frac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}=\frac{2^3\cdot5^2\left(2\cdot5^2+2^3\cdot5\right)}{2^3\cdot5^2}\)
\(=2\cdot5^2+2^3\cdot5=2\cdot25+8\cdot5=50+40=90\)
\(B=\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)
\(B=\frac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}\)
\(B=\frac{2^4\cdot5^3\left(5+2\right)}{2^3\cdot5^2}\)
\(B=\frac{2^4\cdot5^3\cdot7}{2^3\cdot5^2}\)
B = 2 . 5 . 7
B = 70
Không chắc =))
rút gọn biểu thức
B=4^2.25^2+32.125/2^3.5^2
C=3^2.1/243.81^2.1/3^2
D=4^6.256^2.2^4
Các pạn giúp mình na
d: =-2/7-3/11+2/7=-3/11
e: =2+3/7+1+4/7-17/7
=4-17/7=11/7
f: =-2/3*4/5+1/5*11/9
=-8/15+11/45
=-24/45+11/45=-13/45
h: =(-3,1+3,7):0,2=0,6:0,2=3
\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}.\dfrac{5}{1}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{1}{3}.2+\dfrac{3}{5}.\dfrac{1}{3}\)
=\(\dfrac{1}{3}.\left(\dfrac{2}{5}+\dfrac{3}{5}-2\right)\)=\(\dfrac{1}{3}.\left(-1\right)=\dfrac{-1}{3}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{1017.2019}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2017\cdot2019}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\)
\(=\left(\dfrac{9}{11}+\dfrac{20}{11}\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{8}{13}\\ =\dfrac{29}{11}+1+\dfrac{8}{13}\)
Biết đến đó thôi sorry:<
\(\dfrac{4^2.25^2+32.125}{2^3.5^2}\)
\(=\dfrac{\left(2^2\right)^2.\left(5^2\right)^2+2^5.5^3}{2^3.5^2}\)
\(=\dfrac{2^4.5^4+2^5.5^3}{2^3.5^2}\)
\(=\dfrac{2^3.5^2\left(2.5^2+2^2.5\right)}{2^3.5^2}\)
\(=2.5^2+2^2.5\)
\(=2.25+4.5=50+20=70\)