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45^10*5^20/75^15
=5^10*9^10*5^20/(5^2)^15
=5^10*5^20*9^10/5^30
=9^10
(0.8)^5/(0.4)^6
=(0.4)^5*2^5/(0.4)^6
=2^5/(0.4)
=32/(0.4)
=80
2^15*9^4/6^6*8^3
=2^15*(3^2)^4/2^6*3^6*(2^3)^3
=2^15*3^8/2^6*3^6*2^9
=3^2
=9
Bài 1:
ta có: 333<3333; 444<4444
=> 333444<33334444
Bài 2:
\(A=\frac{21^5}{81}=\frac{\left(3.7\right)^5}{3^4}=\frac{3^5.7^5}{3^4}=3.7^5=50421\)
\(B=\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{\left(3.0,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{3^5.\left(0,5\right)^5}=\frac{1}{3^2}=\frac{1}{9}\)
\(C=2^2.\frac{1}{128}.45.2^{-6}=\frac{2^2.45}{128.64}=\frac{2^2.45}{2^7.2^6}=\frac{45}{2^{11}}=\frac{45}{2048}\)
\(D=\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+2^2.3^3+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}\)\(=3^3.\left(-1\right)=-27\)
\(=\dfrac{2}{3}\cdot\dfrac{6}{5}\cdot7+\dfrac{5}{6}\cdot\dfrac{1}{7}\cdot\dfrac{7}{8}\cdot3=\dfrac{28}{5}+\dfrac{5}{16}=\dfrac{473}{80}\)
P=(-0,5-3/5):(-3)+1/3-(-1/6):(-2)
=(-1/2-3/5):(-3)+1/3+1/6:(-2)
= -9/10: -4/3+1/12
= -27/40+1/12
=-71/120
\(A=1.3+2.4+3.5+....+48.50\)
\(A=1.\left(1+2\right)+2.\left(3+1\right)+3.\left(4+1\right)+....+48.\left(49+1\right)\)
\(A=1.2+1+2.3+2+3.4+3+....+48.49+48\)
\(A\left(=1.2+2.3+...+48.49\right)+\left(1+2+...+48\right)\)
tự làm tiếp :))
p/s: ck iu :3
`(6-2/3+1/2)-(5+5/3-3/2)-(3-7/3+5/2)`
`= 6-2/3+1/2 -5-5/3+3/2 -3+7/3 -5/2`
`= (6-5-3) - ( 2/3 - 5/3 + 7/3 ) + (1/2 + 3/2 - 5/2)`
`= -2 -0 + (-1/2)`
`= -5/2`
\(\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=\dfrac{-5}{2}\)
Giải :
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
Ta có :
A= 6 - 5 - 3 - \(\frac{2}{3}\) - \(\frac{5}{3}\) + \(\frac{7}{3}\) + \(\frac{1}{2}\) + \(\frac{3}{2}\) - \(\frac{5}{2}\)
= - 2 - \(\frac{1}{2}\) = \(-\frac{5}{2}\)
\(A=\frac{6^2.6^3}{3^5}\)
\(\Rightarrow A=\frac{6^5}{3^5}\)
\(\Rightarrow A=\frac{7776}{243}\)
\(\Rightarrow A=32\)
tíc mình nha