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Ta có: 76 + 75 - 74
= 74 . (49+7-1)
= 74 . 55 chia hết cho 11 => ĐPCM
Ta có: 2454⋅5424⋅210
= (23 . 3)54 . (33 . 2) . 210
= 2162 . 354 . 372. 224 . 210
= 2196 . 3126
= (2189 . 3126). 27
=7263 . 27 chia hết cho 63 => ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(24^{54}.54^{24}.2^{10}=\left(2^3\right)^{54}.3^{54}.2^{24}.\left(3^3\right)^{24}.2^{10}=2^{196}.3^{126}=2^7.2^{189}.\left(3^2\right)^{63}\)
\(=2^7.\left(2^3\right)^{63}.9^{63}=2^7.8^{63}.9^{63}=2^7.72^{63}\) chia hết cho \(72^{63}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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a. 87 - 218 = 221 - 218 = 217 ( 24 - 2) = 217 ( 16-2) = 217 * 14 chia het cho 14
b. 55 - 54 + 53 = 53 ( 52 - 5 + 1) = 53 * 21 chia het cho 7
con nhung bai lai ban tu giai nhe , con neu thac mac hoi ban
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Bài 1:
a/ \(P\left(x\right)=\frac{1}{2}\left(4x^2+4x+1\right)+\frac{3}{4}=\frac{1}{2}\left(2x+1\right)^2+\frac{3}{4}\)
Do \(\frac{1}{2}\left(2x+1\right)^2\ge0\) \(\forall x\Rightarrow P\left(x\right)=\frac{1}{2}\left(2x+1\right)^2+\frac{3}{4}>0\) \(\forall x\)
\(\Rightarrow\) Đa thức ko có nghiệm
b/ \(72^{63}=\left(8.9\right)^{63}=\left(2^3.3^2\right)^{63}=2^{189}.3^{126}\)
\(A=24^{54}.54^{24}.2^{10}=\left(8.3\right)^{54}.\left(27.2\right)^{24}.2^{10}=\left(2^3.3\right)^{54}.\left(3^3.2\right)^{24}.2^{10}=2^{196}.3^{126}\)
\(\Rightarrow A=2^7.2^{189}.3^{126}=2^7.72^{63}⋮72^{63}\)
Bài 2:
\(5x^2+10x=0\Leftrightarrow5x\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}5x=0\\x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(5^{\left(x-2\right)\left(x+3\right)}=1\Leftrightarrow5^{\left(x-2\right)\left(x+3\right)}=5^0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)