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5 tháng 4 2019
\(\frac{7}{4}-\left(\frac{1}{2.2}+\frac{1}{4.3}+\frac{1}{6.4}+\frac{1}{8.5}+\frac{1}{10.6}+\frac{1}{12.7}+\frac{1}{14.8}\right)\div x=0\)
\((\frac{1}{2.2}+\frac{1}{4.3}+\frac{1}{6.4}+\frac{1}{8.5}+\frac{1}{10.6}+\frac{1}{12.7}+\frac{1}{14.8})\div x=\frac{7}{4}\)
\((\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112})\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\right]\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\right]\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(1-\frac{1}{8}\right)\right]\div x=\frac{7}{4}\)
\(\left(\frac{1}{2}.\frac{7}{8}\right)\div x=\frac{7}{4}\)
\(\frac{7}{16}\div x=\frac{7}{4}\)
\(x=\frac{7}{16}\div\frac{7}{4}\)
\(x=\frac{7}{16}\times\frac{4}{7}\)
\(x=\frac{1}{4}\)
5 tháng 4 2019
\(\frac{7}{4}-\left(\frac{1}{2\cdot2}+\frac{1}{4\cdot3}+\frac{1}{6\cdot4}+\frac{1}{8\cdot5}+\frac{1}{10\cdot6}+\frac{1}{12\cdot7}+\frac{1}{14\cdot8}\right)\)
\(=\frac{7}{4}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(1-\frac{1}{8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\cdot\frac{7}{8}\)
\(=\frac{7}{4}-\frac{7}{16}=\frac{28}{16}-\frac{7}{16}=\frac{21}{16}\)
27 tháng 8 2020
1) Ta có: \(\frac{\left(5^4-5^3\right)^3}{125^4}\)
\(=\frac{\left[5^3\cdot\left(5-1\right)\right]^3}{5^{12}}\)
\(=\frac{5^9\cdot4^3}{5^{12}}\)
\(=\frac{4^3}{5^3}=\frac{64}{125}\)
2) Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+3^{10}\cdot2^{12}\cdot5}{6^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{6^{11}\cdot\left(6-1\right)}\)
\(=\frac{6^{11}\cdot2^2}{6^{11}\cdot5}=\frac{2^2}{5}=\frac{4}{5}\)
VM
25 tháng 1 2020
c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
VM
25 tháng 1 2020
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!
HL
1
19 tháng 3 2017
\(\dfrac{2^{12^{ }}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
= \(\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.14^3}\)
= \(\dfrac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\dfrac{5^{10}\left(7^3-7^4\right)}{5^9\left(7^3+14^3\right)}\)
=\(\dfrac{3^5-3^4}{3^6+3^5}-\dfrac{5\left(7^3-7^4\right)}{7^3+14^3}\)
=\(\dfrac{243-81}{729+243}-\dfrac{5\left(343-2401\right)}{343+2744}\)
=\(\dfrac{162}{972}-\dfrac{5\left(-2058\right)}{3087}\)
=\(\dfrac{1}{6}-\dfrac{-10}{3}\)
= \(\dfrac{1}{6}+\dfrac{20}{6}\)
=\(\dfrac{21}{6}=3,5\)
BB
1
18 tháng 5 2022
1:
\(\Leftrightarrow4\cdot3^x\cdot\dfrac{1}{9}+2\cdot3^x\cdot3=4\cdot3^4+2\cdot3^7\)
\(\Leftrightarrow3^x\cdot\left(\dfrac{4}{9}+6\right)=3^4\cdot\left(4+2\cdot3^3\right)\)
\(\Leftrightarrow3^x=729\)
hay x=6
2: \(\Leftrightarrow3^x\cdot4\cdot\dfrac{1}{3}+3^x\cdot2\cdot9=4\cdot3^6+2\cdot3^9\)
\(\Leftrightarrow3^x\cdot\dfrac{58}{3}=42282\)
=>3x=2187
hay x=7
\(\dfrac{36^4\cdot125}{2^8\cdot5^4\cdot3^6}=\dfrac{2^8\cdot3^8\cdot5^3}{2^8\cdot5^4\cdot3^6}=3^2\cdot\dfrac{1}{5}=\dfrac{9}{5}\)
= ( 2^2 . 3^2)^4 . 5 ^ 3 / 2^8.5^4.3^6
= 2^8 . 3^8 . 5^3 / 2^8 . 5^4 . 3^6
=3^2/5
= 9/5
Chúc bạn học tốt ^^ !!