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ta có B = \(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\left(\dfrac{40}{24.34}+...+\dfrac{40}{54.64}\right)\)
\(=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+...+\dfrac{10}{54.64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+...+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(=5.\dfrac{1}{20}-4.\dfrac{5}{192}\)
\(=5.\dfrac{1}{20}-\dfrac{4}{192}.5\)
\(=5\left(\dfrac{1}{20}-\dfrac{4}{192}\right)=5.\dfrac{7}{240}=\dfrac{7}{48}\)
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)
\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\)
\(A=\frac{1}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+...+\frac{6}{50.56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+...+\frac{1}{50}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\frac{3}{28}\)
\(A=\frac{1}{56}\)
\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)
\(B=5.\left(\frac{9}{12.21}+\frac{9}{21.30}\right)-4.\left(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{64}\right)\)
\(B=5.\frac{1}{20}-4.\frac{5}{192}\)
\(B=\frac{1}{4}-\frac{5}{48}\)
\(B=\frac{7}{48}\)
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 \(\frac{A}{B}< \frac{1}{8}\)
\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+....+\frac{1}{50.56}\)
\(=\frac{1}{6}.(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+....+\frac{6}{50.56})\)
\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+....+\frac{1}{50}-\frac{1}{56})\)
\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{56})\)
\(=\frac{1}{6}.(\frac{7}{56}-\frac{1}{56})\)
\(=\frac{1}{6}.\frac{6}{56}\)
\(=\frac{1}{56}\)
\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)
\(=5(\frac{9}{12.21}+\frac{9}{21.30})-4(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64})\)
\(=5(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64})\)
\(=5(\frac{1}{12}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{64})\)
\(=5(\frac{5}{60}-\frac{2}{60})-(\frac{4}{24}-\frac{4}{64})\)
\(=5.\frac{1}{20}-(\frac{1}{6}-\frac{1}{16})\)
\(=\frac{1}{4}-(\frac{8}{48}-\frac{3}{48})\)
\(=\frac{1}{4}-\frac{5}{48}\)
\(=\frac{12}{48}-\frac{5}{48}=\frac{7}{48}\)
\(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}\)
\(=\frac{1}{56}.\frac{48}{7}\)
\(=\frac{6}{49}=\frac{48}{392}\)bé hơn \(\frac{49}{392}=\frac{1}{8}\)
Vậy \(\frac{A}{B}\)bé hơn \(\frac{1}{8}\)
Chúc bạn học tốt
\(a,\frac{20^{12}\cdot6^{14}}{8^{13}\cdot15^{12}}\)
\(=\frac{5^{12}\cdot2^{24}\cdot2^{14}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}\)
\(=\frac{5^{12}\cdot2^{38}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}=\frac{3^2}{2}=\frac{9}{2}\)
\(b,\frac{45^{12}\cdot10^{14}}{18^{13}\cdot25^{12}}\)
\(=\frac{5^{12}\cdot3^{24}\cdot2^{14}\cdot5^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}\)
\(=\frac{5^{26}\cdot3^{24}\cdot2^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}=\frac{5^2\cdot2}{3^2}=\frac{50}{9}\)
\(c,\frac{18^{12}\cdot27^8}{6^8\cdot3^{40}}\)
\(=\frac{2^{12}\cdot3^{24}\cdot3^{24}}{2^8\cdot3^8\cdot3^{40}}\)
\(=\frac{2^{12}\cdot3^{48}}{2^8\cdot3^{48}}=2^4=16\)
\(d,\frac{12^{14}\cdot9^{18}}{8^9\cdot27^{17}}\)
\(=\frac{3^{14}\cdot2^{28}\cdot3^{36}}{2^{27}\cdot3^{51}}\)
\(=\frac{3^{50}\cdot2^{28}}{2^{27}\cdot3^{51}}=\frac{2}{3}\)
làm hơi tắt nên chịu khó hiểu
\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\Rightarrow\frac{x-9}{y-12}\)
\(\Rightarrow\frac{3}{4}=\frac{x-9}{y-12}=\frac{9}{12}=\frac{x-9}{y-12}=\frac{x-9+9}{y-12+12}\)\(=\frac{x}{y}=\frac{xy}{y^2}=\frac{x^2}{xy}\)
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)
\(\frac{15}{x-9}=\frac{20}{y-12}\Leftrightarrow15\left(y-12\right)=20\left(x-9\right)\Leftrightarrow15y-180=20x-180\Leftrightarrow15y=20x\Leftrightarrow\frac{y}{20}=\frac{x}{15}\Leftrightarrow\frac{xy}{20}=\frac{x^2}{15}\Leftrightarrow x^2=\frac{15.1200}{20}=900\Leftrightarrow\orbr{\begin{cases}x=30\\x=-30\end{cases}}\)
Chia từng trường hợp tìm y, z.
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
Đáp án :\(\frac{65}{192}\)