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a) \(110 – 7^2 + 22:2\) = \(110 – 49 + 11 = 61 + 11 = 72\)
b) \(9.(8^2 – 15)= 9.(64 – 15 ) = 9.49 = 441\)
c) \(5.8 – (17 + 8):5 = 40 – 25:5 = 40 – 5 = 35\)
d) \( 75:3 + 6.9^2= 25 + 6.81 = 25 + 486 = 511\)
a)2/13 . 5 - 9/11 . 2/13 - 7/11 . 2/13
=2/13.(5-9/11-7/11)
=2/13.39/11
=6/11.
b)(-1/2)2 + 4,25
=1/4+4,25
=1/4+17/4
=9/2.
c)(-2/3)2 : ( --2 2/3) -- ( 5/8 -- 5/6 ) + 5.8--45/5.8
= -4/3 : -8/3 - -5/24 + 40 - 72
= -751/24.
Bài 1 :
S = \(\frac{6}{2.5}+\frac{6}{5.8}+...+\frac{6}{29.32}\)
= 2 . \(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
= 2 . \(\left(\frac{1}{2}-\frac{1}{32}\right)\)= ....
\(\text{-7129+1478+7129+(-1479)}\)
\(=\text{-7129+7129+1478+(-1479)}\)
\(=0-1=-1\)
\(\text{|-5|.(-7)+4.(-9)}\)
\(=\text{5.(-7)+4.(-9)}\)
\(=\left(-35\right)+\left(-36\right)=-71\)
Ta có:\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
=>x+3=308
=>x=305
\(\frac{8^{14}}{4^4\cdot64^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
\(\frac{2^7\cdot9^3}{6^5\cdot8^2}=\frac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\frac{3}{2^4}=\frac{3}{16}\)
\(8^5.8^2+8^3=8^7+8^3\)