Tính:
a,A=\(\dfrac{12^{15}.3^4-4^5.3^9}{27^3.2^{10}-32^3.3^9}\)
b. B= \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^3.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{99}{49^2.50^2}\)
Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
Tính
a, A = \(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
b, B = \(\left(\dfrac{3}{7}.\dfrac{4}{15}+\dfrac{1}{3}.9^{15}\right).\dfrac{1}{3}.\dfrac{6^8}{12^4}\)
c, C = \(\dfrac{10^4.81-16.15^2}{4^4.675}\)
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
Bài 1: rút gọn
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}\)
\(\left(\dfrac{1}{5}\right)^{15}\).\(\left(\dfrac{1}{4}\right)^{20}\)
\(\dfrac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
Tính:
a. \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
b. \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
c. \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{25}{12^2.13^2}\)
a: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
b: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\dfrac{2^{10}\cdot3^8\cdot\left(-2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
a) Tìm x biết: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
b) So sánh \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) với 1
c) Tìm GTNN của: A= |x-10|+|x-5|
a/ \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
Mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
\(\Leftrightarrow x+101=0\)
\(\Leftrightarrow x=-101\)
Vậy...
b/ Đặt :
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+.........+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+....+\dfrac{10^2-9^2}{9^2.10^2}\)
\(=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+....+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{10^2}< 1\)
\(\Leftrightarrow A< 1\left(đpcm\right)\)
Vậy...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x-10\right|+\left|x-5\right|=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A\ge\left|x-10+5-x\right|\)
\(\Leftrightarrow A\ge5\)
Dấu "=" xảy ra
\(\Leftrightarrow\left(x-10\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge10\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le10\\5\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\5\le x\le10\end{matrix}\right.\)
Vậy..
Tính
a,\(\dfrac{3}{4}\) - \(\dfrac{1}{12}\) - \(\dfrac{1}{6}\) + 2 b,\(\dfrac{2}{5}\) + \(\dfrac{3}{10}\) + \(\dfrac{4}{5}\) - \(\dfrac{5}{6}\)
c, \(\dfrac{12}{25}\) : \(\dfrac{9}{32}\) \(\dfrac{16}{15}\) d, \(\dfrac{3}{5}\) x \(\dfrac{2}{7}\) x \(\dfrac{5}{6}\) : 5
a: =9/12-1/12-2/12+2=1/2+2=5/2
b: =(2/5+4/5+3/10)-5/6=6/5+3/10-5/6=15/10-5/6=3/2-5/6=9/6-5/6=4/6=2/3
c: \(=\dfrac{12}{25}\cdot\dfrac{32}{9}\cdot\dfrac{15}{16}=\dfrac{12}{9}\cdot\dfrac{32}{16}\cdot\dfrac{15}{25}=\dfrac{4}{3}\cdot\dfrac{3}{5}\cdot2=\dfrac{8}{5}\)
\(\dfrac{-1}{3}.\dfrac{2}{5}\)
\(\dfrac{-3}{7}.\dfrac{4}{15}\)
\(\dfrac{-9}{3}.\dfrac{15}{27}\)
\(\dfrac{-5}{6}.\)( -12 )
\(\dfrac{-12}{4}.\dfrac{8}{9}\)
\(-3.\dfrac{7}{6}\)
\(\dfrac{-15}{9}.\dfrac{27}{5}\)
\(-9.\dfrac{4}{27}\)
-2/15
-4/35
-5/3
10
-8/3
-7/2
-9
-4/3
Chúc em học giỏi
\(=\dfrac{-2}{15}\\ =\dfrac{-4}{35}\\ =-1\\ =10\\ =\dfrac{-8}{3}\\ =-7\\ =-9\\ =\dfrac{-4}{3}\)
Rút gọn rồi tính:
\(\dfrac{3}{12}+\dfrac{1}{4}\) \(\dfrac{4}{10}+\dfrac{3}{5}\) \(\dfrac{12}{27}+\dfrac{2}{9}\) \(\dfrac{7}{3}+\dfrac{20}{15}\)
\(\dfrac{3}{12}+\dfrac{1}{4}=\dfrac{3:3}{12:3}+\dfrac{1}{4}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\dfrac{4}{10}+\dfrac{3}{5}=\dfrac{4:2}{10:2}+\dfrac{3}{5}=\dfrac{2}{5}+\dfrac{3}{5}=\dfrac{5}{5}=1\)
\(\dfrac{12}{27}+\dfrac{2}{9}=\dfrac{12:3}{27:3}+\dfrac{2}{9}=\dfrac{4}{9}+\dfrac{2}{9}=\dfrac{6}{9}=\dfrac{2}{3}\)
\(\dfrac{7}{3}+\dfrac{20}{15}=\dfrac{7}{3}+\dfrac{20:5}{15:5}=\dfrac{7}{3}+\dfrac{4}{3}=\dfrac{11}{3}\)
\(\dfrac{2^4.5^2.7}{2^3.5.7^2.11}\);\(\dfrac{2^3.3^3.5^3.7.8}{3.2^4.5^3.14}\)
\(\dfrac{4^2.5.11}{44.20}\);\(\dfrac{13.15.16}{18.65.7};\dfrac{7.2.8.5^2}{14.2.5}\)
\(\dfrac{2^3.3^3.5}{3.2^3.5^3}\)
a) \(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2^3\cdot5\cdot10\cdot7}{2^3\cdot5\cdot7\cdot77}=\dfrac{10}{77}\)
\(\dfrac{2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot2^4\cdot5^3\cdot14}=\dfrac{2^3\cdot3\cdot5^3\cdot7\cdot3^2\cdot8}{3\cdot2^3\cdot2\cdot5^3\cdot14}=\dfrac{7\cdot3^2\cdot8}{2\cdot14}=\dfrac{63\cdot8}{2\cdot14}=18=\dfrac{1386}{77}\)