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LB
1
22 tháng 9 2017
\(A=\dfrac{\left(12.8\right)^3.5}{2^9.15^2}=\dfrac{\left(2^2.3.2^3\right)^3.5}{2^9.\left(3.5\right)^2}=\dfrac{2^6.3^3.2^9.5}{2^9.3^2.5^2}=\dfrac{2^{15}.3^3.5}{2^9.3^2.5^2}=\dfrac{2^6.3.1}{1.1.5}=\dfrac{192}{5}\)
21 tháng 12 2017
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+.........+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+..........+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
\(\Leftrightarrow2A=1-\dfrac{1}{2n+1}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{2n+1}\right).\dfrac{1}{2}\)
\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2n+1}< \dfrac{1}{2}\)
\(\Leftrightarrow A< \dfrac{1}{2}\left(đpcm\right)\)
28 tháng 3 2017
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
DH
13 tháng 7 2017
a,?????
b, Với mọi giá trị của x;y ta có:
\(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|\ge0\)
Để \(\left|x-\dfrac{1}{2}\right|+\left|x+y\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{2}+y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..........
c, \(\left|2x\right|-\left|3,5\right|=\left|-6,5\right|\)
\(\Rightarrow\left|2x\right|=6,5+3,5=10\)
\(\Rightarrow\left\{{}\begin{matrix}2x=10\\2x=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy..........
d, \(\left|x-1,7\right|=2,3\)
\(\Rightarrow\left\{{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!
FC
18 tháng 10 2018
a, (5-5)-1 . \(\left(\dfrac{1}{2}\right)^{-2}\) . \(\dfrac{1}{10^5}\)
= 55 . \(\dfrac{1^{-2}}{2^{-2}}\) . \(\dfrac{1}{10^5}\)
= (55 . \(\dfrac{1}{10^5}\)) . \(\dfrac{1}{4}\)
= \(\dfrac{5^5}{10^5}\) . \(\dfrac{1}{4}\) = \(\left(\dfrac{1}{2}\right)^5\). \(\dfrac{1}{4}\)
= \(\dfrac{1}{32}.\text{}\dfrac{1}{4}\)= \(\dfrac{1}{128}\)
21 tháng 10 2022
b: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{4}{5}\)
c: \(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^3\cdot5^2}=10\cdot7=70\)
DV
8 tháng 12 2017
\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
\(=1-\dfrac{1}{2n+1}\Rightarrow A=\left(1-\dfrac{1}{2n+1}\right)\cdot\dfrac{1}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2n+1}< \dfrac{1}{2}\)
Vậy A < \(\dfrac{1}{2}\)
17 tháng 8 2018
a, \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\Leftrightarrow98\left(2x+1\right)=99.2x\)
\(\Leftrightarrow2x=98\Rightarrow x=49\)
17 tháng 6 2022
b: Đặt \(A=1-3+3^2-3^3+...+\left(-3\right)^x\)
\(=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^x\)
\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)
\(\Leftrightarrow-3A-A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}-...-1\)
\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)
\(\Leftrightarrow A=\dfrac{\left(-3\right)^{x+1}-1}{-4}=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)
\(\Leftrightarrow\dfrac{-\left(-3\right)^{x+1}+1}{4}=\dfrac{3^{2012}-1}{2}\)
\(\Leftrightarrow-\left(-3\right)^{x+1}+1=2\cdot3^{2012}-2\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=2\cdot3^{2012}-3\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=3\left(2\cdot3^{2011}-1\right)\)
\(\Leftrightarrow-\left(-3\right)^x=2\cdot3^{2011}-1\)
=>x=2010
30 tháng 5 2022
a: \(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\cdot12=20\)
b: \(=\dfrac{3}{5}:\left(\dfrac{-2-5}{30}\right)+\dfrac{3}{5}:\left(\dfrac{-1}{3}-\dfrac{16}{15}\right)\)
\(=\dfrac{3}{5}:\dfrac{-7}{30}+\dfrac{3}{5}:\dfrac{-21}{15}\)
\(=\dfrac{3}{5}\left(\dfrac{-30}{7}-\dfrac{15}{21}\right)=\dfrac{3}{5}\cdot\left(\dfrac{-30}{7}-\dfrac{5}{7}\right)=\dfrac{3}{5}\cdot\left(-5\right)=-3\)
c: \(=5.7\left(-6.5-3.5\right)\)
\(=5.7\cdot\left(-10\right)=-57\)
d: \(=10\cdot0.1\cdot\dfrac{4}{3}+3\cdot7-\dfrac{1}{6}\cdot2\)
\(=\dfrac{4}{3}+21-\dfrac{1}{3}=22\)
A=\(\dfrac{12^3\cdot8^3\cdot5}{2^9\cdot\left(3\cdot5\right)^2}\)
A=\(\dfrac{\left(2^2\cdot3\right)^3\cdot\left(2^3\right)^3\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{\left(2^2\right)^3\cdot3^3\cdot2^9\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{2^6\cdot3^3\cdot2^9\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{2^6\cdot3}{5}\)
A=\(\dfrac{64\cdot3}{5}\)
A=\(\dfrac{192}{5}\)
Vậy A=\(\dfrac{192}{5}\)
Tích cho mình nhé!