a, 1 x 2 x 3 x ... x 8 x 9 - 1 x 2 x 3 x ... x 8 - 1 x 2 x 3 x ... x 7 x 8 x 8 

= 1 x 2 x 3 x ... x 8 ( 9 - 1 - 8 ) = 1 x2 x 3 x ... x 8 . 0 = 0 

Bài post của vuong hien duc

a, 1.2.3...9-1.2.3...8-1.2.3...7.8²

= 1.2.3...7.8(9-1-8) = 0

b,\(\frac{3.4.2^{16}}{11.2^{13}.4^{11}-16^9}=\frac{3.2^2.2^{16}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\) \(=\frac{3.2^{18}}{11.2^{13}.2^{22}-2^{36}}=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)\(=\frac{3.2^{18}}{2^{35}\left(11-2\right)}=\frac{3.2^{18}}{2^{35}.9}=\frac{1}{2^{17}.3}\)

\(B=\frac{(3\cdot4\cdot2^{16})}{11\cdot2^{13}\cdot4^{11}-16^9}\)

\(=\frac{(2^2)^2\cdot(2^{16})^2\cdot3^2}{2^{13}\cdot(2^2)^{11}\cdot11-(2^4)^9}\)

\(=\frac{2^4\cdot2^{32}\cdot3^2}{2^{13}\cdot2^{22}\cdot11-2^{36}}\)

\(=\frac{2^{36}\cdot3^2}{2^{35}\cdot11-2^{36}}\)

\(=\frac{2^{36}\cdot3^2}{2^{35}\cdot(11-2)}\)

\(=\frac{2^1\cdot3^2}{9}\)

\(=\frac{2\cdot3^2}{3^2}=2\)

\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)

\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)

\(=\frac{2^9.3^9.-17}{1}\)

Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)

\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)

\(=\frac{3.2^{18}}{2^{35}.3^2}\)

\(=\frac{1}{2^{17}.3}\)

Cảm ơn bạn nhé:)

H = 3^2.4^4.2^32/11.2^13.(2^2)^11-(2^4)^9

   = 3^2.(2^2)^4.2^32/11.2^13.2^22-2^36

   = 3^2.2^8.2^32/11.2^35-2^36

   = 3^2.2^40/2^35.(11-2)

   = 9.2^40/2^35.9 = 2^5 = 32

k mk nha

a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)

\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\cdot\frac{8}{33}\)

\(=\frac{52}{33}\)

a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99

A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)

A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)

A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)

A= 13/2 ( 1/3 - 1/11) 

A= 13/2 . 8/33

A= 52/33  

\(b,\)\(\left(\frac{15}{1.2.3}+\frac{15}{2.3.4}+\frac{15}{3.4.5}+...+\frac{15}{18.19.20}\right).x=1\)

\(\left[\frac{15}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{15}{18.19.20}\right)\right].x=1\)

\(\left[\frac{15}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right].x=1\)

\(\left[\frac{15}{2}.\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\right].x=1\)

\(\left[\frac{15}{2}.\frac{189}{380}\right].x=1\)

\(\frac{567}{152}.x=1\)

\(x=1-\frac{567}{152}\)

\(\Rightarrow x=-\frac{415}{152}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)  vậy

\(\frac{11}{45}.x=\frac{22}{45}\)

\(x=\frac{22}{45}\div\frac{11}{45}=2\)

vậy suy ra x =2

mình chắc chắn 100% luôn đó, cái này ở trong violympic toán 7 vòng 12 phải ko
 

X=2 nhé bạn.....đúng đó, vòng 12 mk 300 mà cx gặp câu này!!! Tick nha

3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)

\(\Leftrightarrow2x-6-3+6x=4+4-4x\)

\(\Leftrightarrow8x-9=8-4x\)

\(\Leftrightarrow8x+4x=8+9\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\dfrac{17}{12}\)

Vậy \(x=\dfrac{17}{12}\)

4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)

\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)

\(\Leftrightarrow6x-12-4-4x=12-9x-12\)

\(\Leftrightarrow6x-4-4x=12-9x\)

\(\Leftrightarrow2x-4=12-9x\)

\(\Leftrightarrow2x+9x=12+4\)

\(\Leftrightarrow11x=16\)

\(\Leftrightarrow x=\dfrac{16}{11}\)

Vậy \(x=\dfrac{16}{11}\)

giúp em câu c) với ạ

c) \(\frac{\left(3\cdot4\cdot2^{16}\right)}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot2^{22}-2^{36}}\)

\(=\frac{9\cdot2^4\cdot2^{32}}{11\cdot2^{35}-2^{26}}\)

\(=\frac{9\cdot2^4\cdot2^{32}2^{ }}{\left(11-2\right)\cdot2^{35}}\)

\(=\frac{9\cdot2^4\cdot2^{32}}{9\cdot2^{35}}\)

\(=\frac{9\cdot1\cdot2^{32}}{9\cdot2^{31}}=\frac{2^{32}}{2^{31}}=2\)