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Sửa đề: 3^5+3^5+3^5; 2^x
=>\(2^x=\dfrac{4^5\cdot4}{3^5\cdot3}\cdot\dfrac{6^5\cdot6}{2^5\cdot2}\)
=>\(2^x=\left(\dfrac{4}{3}\right)^6\cdot\left(\dfrac{6}{2}\right)^6=4^6=2^{12}\)
=>x=12
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\)
\(\Rightarrow\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\)
\(\Rightarrow\dfrac{4^5.4.6^5.6}{3^5.3.2^5.2}=2^n\)
\(\Rightarrow\dfrac{\left(2.2\right)^5.2.2.\left(3.2\right)^5.3.2}{3^5.3.2^5.2}=2^n\)
\(\Rightarrow\dfrac{2^5.2^5.2.2.3^5.2^5.3.2}{3^5.3.2^5.2}=2^n\)
Rút gọn vế trái ta có :
\(2^5.2.2.^5=2^n\)
\(\Rightarrow2^{12}=2^n\)
\(\Rightarrow n=12\) ( Thỏa mãn điều kiện \(n\in N\) )
Vậy n =12
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=>\(\dfrac{4^5\left(1+1+1+1\right)}{3^5\left(1+1+1\right)}.\dfrac{6^5\left(1+1+1+1+1+1\right)}{2^5\left(1+1\right)}=2^n\)
=>\(\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) =>\(\dfrac{4^6}{3^6}.\dfrac{6^6}{2^6}=2^n\)
=>\(\left(\dfrac{4.6}{3.2}\right)^6=2^n\) =>\(4^6=2^n\) =>\(2^{12}=2^n\) =>n=12.
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1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\)*\(\frac{6^5+6^5+^{_{6^5}}+6^5+6^5+6^5}{2^5+2^5}\)=\(^{2^x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)
\(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}\)
\(=\frac{4^6}{3^6}.\frac{6^6}{2^6}\)
\(=\frac{4^6.\left(2.3\right)^6}{\left(2.3\right)^6}\)
\(=4^6=\left(2^2\right)^6\)
\(=2^{2.6}=2^{12}\)
\(\Rightarrow n=12\)
Vậy n=12
trong cái xã hội này có làm thì mới có ăn,ko lam mà ăn chỉ có ăn đầu b** ăn c**
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{4\cdot4^5}{3\cdot3^5}\cdot\frac{6\cdot6^5}{2\cdot2^5}=\frac{4^6}{3^6}\cdot\frac{6^6}{2^6}=\frac{2^{12}\cdot2^6\cdot3^6}{3^6\cdot2^6}=2^{12}=2^n\Rightarrow n=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)
\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)
Làm nốt.
ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.
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5: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=3-5-6+\dfrac{-1}{4}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{6}{5}-\dfrac{3}{2}\)
\(=-8+\dfrac{3}{2}+1+\dfrac{-3}{10}\)
\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)
6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=6-5-3-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}+\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)
\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)
=-3+1
=-2
8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)
\(=11+2-1-\dfrac{1}{2}\)
=11+1/2
=11,5
\(\left(5^6\right)^2=5^{12}\)
(5^6)^2
= 5^6.2
= 5^12
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