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\(\left(64.27\right)^6=4^{3.6}.3^{3.6}=\left(4.3\right)^{18}=12^{18}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(64.27\right)^6.75^{18}=\left(2^6.3^3\right)^6.\left(3.5^2\right)^{18}=2^{36}.3^{36}.5^{36}=\left(2.3.5\right)^{36}=30^{36}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: `A` lớn nhất `<=> (2015)/(18+12|x-6|)` nhỏ nhất.
`<=> 18+12|x-6|` nhỏ nhất.
`<=> 12|x-6|` nhỏ nhất, do `18` là hằng.
`<=> 12|x-6|=0`
`<=> x=6 => A=2015/18`
Vậy...
`b, B>=x+1/3+1-x`
`=4/3`.
Đẳng thức xảy ra `<=> x+1/3=1-x`
`<=> x=2/3`.
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
Chia tất cả cho 30=>
\(\frac{x+y}{5}=\frac{y+z}{10}=\frac{x+z}{6}=\frac{x+z-x-y}{6-5}=\frac{y+z-x-z}{10-6}=\frac{z-y}{1}=\frac{y-x}{4}\)
\(\Leftrightarrow\frac{z-y}{3}=\frac{y-x}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(VT=12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
\(VP=18^{16}=2^{16}\cdot3^{32}\)
=> VT=VP
b) \(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
(đề sai)
c) \(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
\(VT=\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left[3^3\left(3-1\right)\right]^2}=\frac{1}{2^2}=\frac{1}{4}=VP\)
Tính nhanh
\(6\left(\frac{-2}{3}\right)+12\left(\frac{-2}{3}\right)^2+18\left(\frac{-2}{3}\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6\left(\frac{-2}{3}\right)+12\left(\frac{-2}{3}\right)^2+18\left(\frac{-2}{3}\right)^2\)
\(=\frac{-2}{3}\times6\times\left[1+2\left(\frac{-2}{3}\right)+3\left(\frac{-2}{3}\right)\right]\)
\(=-4\times\left[1+\left(\frac{-2}{3}\right)\left(2+3\right)\right]\)
\(=-4\times\left(1+\frac{-10}{3}\right)\)
\(=-4+\frac{40}{3}\)
\(=-\frac{12}{3}+\frac{40}{3}\)
\(=\frac{28}{3}\)
\(\left(64.27\right)^6=12^{18}\Leftrightarrow1728^6=12^8\)
Ta có :
\(12^{18}=\left(12^3\right)^6=1728^6\)
Vì \(1728^6=1728^6\)
\(\Leftrightarrow\left(64.27\right)^6=12^{18}\) đó là điều mà ta phải chứng minh