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\(8^{15}=\left(2^3\right)^{15}=2^{3.15}=2^{45}\\ 16^4=\left(2^4\right)^4=2^{4.4}=2^{16}\\ 2^{45}>2^{16}\Rightarrow8^{15}>16^4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
`x^2 -144=0`
`<=> x^2 =144`
\(< =>\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)
2)`
`2x^2 -72=0`
`<=>2x^2 =72`
`<=>x^2=36`
\(< =>\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
3)
`5x^2 -125=0`
`<=> 5x^2 =125`
`<=>x^2 =25`
\(< =>\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
4)
`-x^2 +81=0`
`<=> x^2 -81=0`
`<=> x^2 =81`
\(< =>\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
5)
`x(2x-18)=0`
\(< =>\left[{}\begin{matrix}x=0\\2x-18=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
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− 4 5 ⋅ 3 2 16 = − 4 5 . 50 16 = − 4.50 5.16 = − 10 4 = − 5 2 = − 2 1 2
![](https://rs.olm.vn/images/avt/0.png?1311)
ta co :
3215/3216<1
3104/3103>1
vi 3215/3216<1<3104/3103 nen3215/3216<3104/3103
3215/3216 <1 vì 3215 <3216
3104/3103 > 1 vì 3104>3103
Do đó 3104/3103 > 3215/3216
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Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{20}=\dfrac{b}{12}=\dfrac{c}{9}=\dfrac{a+b+c}{20+12+9}=\dfrac{164}{41}=4\)
Do đó: a=80; b=48; c=36
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(3^4\right)^5.\left(2^5\right)^4=3^{20}.2^{20}=6^{20}\)
Kết quả làm tròn số 89, 3216 đến số thập phân thứ 3 là:
A. 89, 321. B. 89, 322. C. 89, 324 D. 89, 326
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`C=(12/35-6/7+18/14):6/(-7)-(-2)/5-1`
`C=(12/35-6/7+9/7):(-6)/7+2/5-1`
`C=(12/35+3/7):(-6)/7+2/5-1`
`C=(12/35+15/35):(-6)/7+2/5-1`
`C=27/35*(-7)/6+2/5-1`
`C=-9/10+2/5-1`
`C=-9/10+4/10-1`
`C=-1/2-1=-3/2`
Ta có: \(C=\left(\dfrac{12}{35}-\dfrac{6}{7}+\dfrac{18}{14}\right):\dfrac{6}{-7}-\dfrac{-2}{5}-1\)
\(=\left(\dfrac{24}{70}-\dfrac{60}{70}+\dfrac{90}{70}\right):\dfrac{-6}{7}-\dfrac{-2}{5}-\dfrac{5}{5}\)
\(=\dfrac{27}{35}\cdot\dfrac{-7}{6}-\dfrac{3}{5}\)
\(=\dfrac{-3}{2}\)
\(\frac{\left(-2\right)^{18}.8^{15}.16^4}{32^{16}}\)
\(=\frac{2^{18}.2^{45}.2^{16}}{2^{80}}\)
\(=\frac{2^{79}}{2^{80}}=\frac{1}{2}\)
(-2)^18.8^15.16^4/32^16
= 1/2
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