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.

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) \( 75:3 + 6.9^2= 25 + 6.81 = 25 + 486 = 511\)

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

Nhân 3 vào cả 2 vế rồi tách ra là làm đc thui!

a: ta có: \(5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5\)

\(=5\cdot\left(2^3\right)^{12}\cdot\left(3^2\right)^5\cdot3^2\cdot\left(-1\right)\cdot2^{20}\)

\(=-5\cdot3^{12}\cdot2^{56}\)

Ta có: \(7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5\)

\(=-7\cdot\left(7^2\right)^3\cdot\left(3^4\right)^5\cdot7^{10}\cdot\left(2^3\right)^5\)

\(=-7\cdot7^6\cdot3^{20}\cdot7^{10}\cdot2^{15}=-7^{17}\cdot2^{15}\cdot3^{20}\)

Ta có: \(B=\frac{5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5}{7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5}\)

\(=\frac{-5\cdot3^{12}\cdot2^{56}}{-7^{17}\cdot2^{15}\cdot3^{20}}=\frac{5\cdot2^{41}}{7^{17}\cdot3^8}\)

b: \(13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4\)

\(=13\cdot2^{14}\cdot3^{20}-3^{14}\cdot2^{16}\)

\(=3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)\)

\(15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}\)

\(=3\cdot5\cdot3^{12}\cdot2^8+3\cdot\left(3^3\right)^5\cdot\left(2^3\right)^{10}=3^{13}\cdot5\cdot2^8+3^{16}\cdot2^{30}\)

\(=3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)\)
Ta có: \(C=\frac{13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4}{15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}}\)

\(=\frac{3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)}{3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)}=\frac{3\cdot2^6\cdot\left(13\cdot3^6-2^2\right)}{5+3^3\cdot2^{22}}\)

\(\frac{8^{14}}{4^4\cdot64^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)

\(\frac{2^7\cdot9^3}{6^5\cdot8^2}=\frac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\frac{3}{2^4}=\frac{3}{16}\)