K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

18 tháng 9 2018

B=5(1/12−1/21+1/21−1/30)−5(1/24−1/34+1/34−1/44+1/44−1/54+1/54−1/64)

B=5(1/12−1/21+1/21−1/30+1/24−1/34+1/34−1/44+1/44−1/54+1/54−1/64 )

B=5(1/12−1/64)=5.13/192=65/192

18 tháng 9 2018

Đáp án :\(\frac{65}{192}\)

18 tháng 9 2018

ta có B = \(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\left(\dfrac{40}{24.34}+...+\dfrac{40}{54.64}\right)\)

\(=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+...+\dfrac{10}{54.64}\right)\)

\(=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+...+\dfrac{1}{54}-\dfrac{1}{64}\right)\)

\(=5\left(\dfrac{1}{12}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)

\(=5.\dfrac{1}{20}-4.\dfrac{5}{192}\)

\(=5.\dfrac{1}{20}-\dfrac{4}{192}.5\)

\(=5\left(\dfrac{1}{20}-\dfrac{4}{192}\right)=5.\dfrac{7}{240}=\dfrac{7}{48}\)

15 tháng 7 2020

\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\)

\(A=\frac{1}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+...+\frac{6}{50.56}\right)\)

\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+...+\frac{1}{50}-\frac{1}{56}\right)\)

\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{56}\right)\)

\(A=\frac{1}{6}.\frac{3}{28}\)

\(A=\frac{1}{56}\)

\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)

\(B=5.\left(\frac{9}{12.21}+\frac{9}{21.30}\right)-4.\left(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64}\right)\)

\(B=5.\left(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64}\right)\)

\(B=5.\left(\frac{1}{12}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{64}\right)\)

\(B=5.\frac{1}{20}-4.\frac{5}{192}\)

\(B=\frac{1}{4}-\frac{5}{48}\)

\(B=\frac{7}{48}\)

Ta có \(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}=\frac{1}{56}\times\frac{48}{7}=\frac{6}{49}\)

Lấy \(\frac{6}{49}-\frac{1}{8}=-\frac{1}{392}< 0\)

\(\Rightarrow\frac{6}{49}< \frac{1}{8}\) hay \(\frac{A}{B}< \frac{1}{8}\)

15 tháng 7 2020

\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+....+\frac{1}{50.56}\)

\(=\frac{1}{6}.(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+....+\frac{6}{50.56})\)

\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+....+\frac{1}{50}-\frac{1}{56})\)

\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{56})\)

\(=\frac{1}{6}.(\frac{7}{56}-\frac{1}{56})\)

\(=\frac{1}{6}.\frac{6}{56}\)

\(=\frac{1}{56}\)

\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)

\(=5(\frac{9}{12.21}+\frac{9}{21.30})-4(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64})\)

\(=5(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64})\)

\(=5(\frac{1}{12}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{64})\)

\(=5(\frac{5}{60}-\frac{2}{60})-(\frac{4}{24}-\frac{4}{64})\)

\(=5.\frac{1}{20}-(\frac{1}{6}-\frac{1}{16})\)

\(=\frac{1}{4}-(\frac{8}{48}-\frac{3}{48})\)

\(=\frac{1}{4}-\frac{5}{48}\)

\(=\frac{12}{48}-\frac{5}{48}=\frac{7}{48}\)

\(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}\)

\(=\frac{1}{56}.\frac{48}{7}\)

\(=\frac{6}{49}=\frac{48}{392}\)bé hơn \(\frac{49}{392}=\frac{1}{8}\)

Vậy \(\frac{A}{B}\)bé hơn \(\frac{1}{8}\)

Chúc bạn học tốt

26 tháng 10 2019

\(a,\frac{20^{12}\cdot6^{14}}{8^{13}\cdot15^{12}}\)

\(=\frac{5^{12}\cdot2^{24}\cdot2^{14}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}\)

\(=\frac{5^{12}\cdot2^{38}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}=\frac{3^2}{2}=\frac{9}{2}\)

\(b,\frac{45^{12}\cdot10^{14}}{18^{13}\cdot25^{12}}\)

\(=\frac{5^{12}\cdot3^{24}\cdot2^{14}\cdot5^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}\)

\(=\frac{5^{26}\cdot3^{24}\cdot2^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}=\frac{5^2\cdot2}{3^2}=\frac{50}{9}\)

\(c,\frac{18^{12}\cdot27^8}{6^8\cdot3^{40}}\)

\(=\frac{2^{12}\cdot3^{24}\cdot3^{24}}{2^8\cdot3^8\cdot3^{40}}\)

\(=\frac{2^{12}\cdot3^{48}}{2^8\cdot3^{48}}=2^4=16\)

\(d,\frac{12^{14}\cdot9^{18}}{8^9\cdot27^{17}}\)

\(=\frac{3^{14}\cdot2^{28}\cdot3^{36}}{2^{27}\cdot3^{51}}\)

\(=\frac{3^{50}\cdot2^{28}}{2^{27}\cdot3^{51}}=\frac{2}{3}\)

làm hơi tắt nên chịu khó hiểu

9 tháng 11 2016

Ta có: \(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\)

\(\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\)

\(\Rightarrow\frac{x}{15}-\frac{9}{15}=\frac{y}{20}-\frac{12}{20}=\frac{z}{40}-\frac{24}{40}\)

\(=\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\)

\(\Rightarrow\frac{x^2}{15^2}=\frac{y^2}{20^2}=\frac{z^2}{40^2}=\frac{xy}{15.20}=\frac{1200}{300}=4=2^2\)

\(\Rightarrow\begin{cases}x^2=2^2.15^2=30^2\\y^2=2^2.20^2=40^2\\z^2=2^2.40^2=80^2\end{cases}\)\(\Rightarrow\begin{cases}x\in\left\{30;-30\right\}\\y\in\left\{40;-40\right\}\\z\in\left\{80;-80\right\}\end{cases}\)

Vậy giá trị (x;y;z) tương ứng thỏa mãn là: (30;40;80) ; (-30;-40;-80)

9 tháng 11 2016

soyeon_Tiểubàng giải thanks bn nhìu nhá!yeu

5 tháng 7 2016

\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{y-12}\)

\(\Rightarrow\frac{3}{4}=\frac{x-9}{y-12}=\frac{9}{12}=\frac{x-9}{y-12}=\frac{x-9+9}{y-12+12}\)\(=\frac{x}{y}=\frac{xy}{y^2}=\frac{x^2}{xy}\)

Từ \(\frac{3}{4}=\frac{xy}{y^2}\Rightarrow\frac{3}{4}=\frac{1200}{y^2}\Rightarrow y^2=1200\cdot\frac{4}{3}=20^2\Rightarrow y=\pm40\)

  • Nếu y=40 => x= 1200: 40 = 30 

Mà \(\frac{15}{x-9}=\frac{40}{z-24}\Rightarrow z=80\)

  • Nếu y = -40 => x = 1200:(-40) = - 30 

Mà \(\frac{15}{x-9}=\frac{40}{z-24}\Rightarrow z=-80\)

Vây (x , y , z ) = ( 30, 40, 80); ( - 30; -40; -80) 

31 tháng 7 2016

ai giúp mk vs

31 tháng 7 2016

Câu a và câu  b khó quá nên minh chí giúp bn câu b thôi!vui

 

31 tháng 7 2016

 

\(\Rightarrow\left[\begin{array}{nghiempt}x-9=15k\\y-12=20k\\z-24=40k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=15k+9\\y=20k+12\\z=40k+24\end{array}\right.}\)

ta có:

x.y=1200\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)

=> (15k+9)(20k+12)=1200

=> 3.4(5k+3)(5k+3)=1200

=> (5k+3)2=100

=> 5k+3=\(\pm\)10

 => \(\left[\begin{array}{nghiempt}5k+3=10\\5k+3=-10\end{cases}\Rightarrow\left[\begin{array}{nghiempt}5k=7\\5k=-13\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}k=\frac{7}{5}\\k=-\frac{13}{5}\end{array}\right.}\)

* với k=7/5

x=7/5x15+9=30

y=7/5x20+12=40

z=7/5x40+24=80

* với k=-13/5

x=-13/5x15+9=-30

y=-13/5x20+12=-40

z=-13/5x40+24=-80

b)

\(\frac{40}{x-30}=\frac{20}{y-50}=\frac{28}{z-21}\Rightarrow\frac{x-30}{40}=\frac{y-50}{20}=\frac{z-21}{28}k=\)

=>\(\left[\begin{array}{nghiempt}x-30=40k\\y-50=20k\\z-21=28k\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=40k+30\\y=20k+50\\z=28k+21\end{array}\right.}\)

ta có:

x.y.z=22400

=> (40k+30)(20k+50)(28k+21)=22400

 

c) 15x=-10y=6z

\(\Rightarrow\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\Rightarrow\frac{x}{2}=-\frac{y}{3}=\frac{z}{5}=k\)

=> \(\left[\begin{array}{nghiempt}x=2k\\y=-3k\\z=5k\end{array}\right.\)

ta có:

x.y.z=30000

=> 2k.(-3k).5k=30000

=> k3=1000

=> k=10

ta có: x=10x2=20

          y=10.(-3)=-30

          z=10.5=50

31 tháng 7 2016

giúp mk vs ! mai đi hok ùi