`8/5 .2/3 + (-5.5)/(3.5) = 16/15 - 5/3 = -3/5`

b) 6/7+5/8 :5 -3/16 .(-2)^2=6/7 + 1/8 - 3/16 .4`

`=55/56 - 3/4`

`=13/56`

\(a,\dfrac{8}{5}.\dfrac{2}{3}+\dfrac{-5.5}{3.5}=\dfrac{16}{15}+\dfrac{-25}{15}=-\dfrac{9}{15}=-\dfrac{3}{5}\)

\(b,\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{55}{56}-\dfrac{3}{4}=\dfrac{13}{56}\)

`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`

`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`

`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`

`=(1-1/7).[130-4-125]/3`

`=6/7 . 1/3 = 2/7`

____________________________________________________

`8/9+1/9 . 2/9+1/9 . 7/9`

`=8/9+1/9.(2/9+7/9)`

`=8/9+1/9 . 9/9`

`=8/9+1/9=9/9=1`

a) ^{_{\(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)}}

^{_{\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)}}

^{_{\(=-\dfrac{5}{24}\)}}

b) ^{_{\(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)}}

^{_{\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)}}

_{^{\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{2}{7}\)}}

_{^{\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)}}

_{^{\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_{^{\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_^\(=1\)

B=(2/5)⁷.5⁷+(9/4)³:(3/16)³/2⁷.5²+512

B=(2/5.5)⁷+(9/4:3/16)³/2⁷.25+512

B=2⁷+12³/2⁷.25+512

B=1856/3200+512

B=1856/3712

B=1/2

\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

\(B=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7\cdot5^2+2^9}\)

\(=\dfrac{1+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{\left(12+1\right)\left(12^2-12+1\right)}{2^7\cdot29}\)

\(=\dfrac{13\cdot133}{2^7\cdot29}=\dfrac{1729}{3712}\)

`# \text {Ryo}`

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\\ =\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\\ =\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-\dfrac{1}{14}\\ =\dfrac{1}{2}-\dfrac{1}{14}\\ =\dfrac{7}{14}-\dfrac{1}{14}\\ =\dfrac{6}{14}\\ =\dfrac{3}{7}\)

3/7

a: \(=\dfrac{2^{13}\cdot5^7\left(2^{17}+5^{20}\right)}{2^{10}\cdot5^7\left(2^{17}+5^{20}\right)}=2^3\)

b: \(M=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)

\(=3^{2\cdot1\cdot13\cdot12}=3^{312}\)

\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)

Đặt A=\(\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)

A=\(\dfrac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}\)

A=\(\dfrac{2^3.5^3.7^4}{2^2.5^2.7^4}\)

A=2.\(5^2\)

A=2.25

A=50