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Câu b:
Ta có: \(x^2 + 4y^2 + z^2 - 2x - 6z + 8y + 15\)
\(= (x^2 - 2x +1) + (4y^2 - 8y + 4) + (z^2 - 6z +9) +1\)
\(= (x-1)^2 + (2y-2)^2 + (z-3)^2 + 1\)
Mà \((x-1)^2 \geq 0; (2y-2)^2 \geq 0; (z-3)^2\geq 0\)
\(\implies\) \((x-1)^2+(2y-2)^2 +(z-3)^2\geq 0\)
\(\implies\)\((x-1)^2+(2y-2)^2 +(z-3)^2+1> 0\)
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\(\left\{{}\begin{matrix}f\left(0\right)⋮5\Rightarrow c⋮5\\f\left(1\right)⋮5\Rightarrow\left(a+b+c\right)⋮5\\f\left(-1\right)⋮5\Rightarrow\left(a-b+c\right)⋮5\\\left[\left(a+b+c\right)+\left(a-b+c\right)\right]=2\left(a+c\right)⋮5\Rightarrow a⋮5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c⋮5\\a⋮5\\b⋮5\end{matrix}\right.\)+> dpcm
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Ta có : \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)\)
\(=a\left(a^2-1\right)\left(a^2-4+5\right)=a\left(a-1\right)\left(a+1\right)\left[\left(a-2\right)\left(a+2\right)+5\right]\)
\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a^2-1\right)\)
Đến đây bạn lập luận đi !
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a/ Đặt \(x^{10}=a\) ta có:
\(A=a^{197}+a^{193}+a^{198}\)
\(=a^{193}\left(a^4+1+a^5\right)\)
\(=a^{193}\left[\left(a^5+a^4+a^3\right)-\left(a^3+a^2+a\right)+\left(a^2+a+1\right)\right]\)
\(=a^{193}\left(a^2+a+1\right)\left(a^3-a+1\right)⋮\left(a^2+a+1\right)\)
Vậy có ĐPCM
b/ \(B=7.5^{2n}+12.6^n=\left(7.25^n-7.6^n\right)+19.6^n\)
\(=7\left(25-6\right)G\left(n\right)+19.6^n=7.19.G\left(n\right)+19.6^n⋮19\)
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chỉ cần CM \(Q=2^{2^n}+4^n+1⋮3\) là ok
Với n=1 thì \(Q⋮3\)
Giả sử Q vẫn chia hết cho 3 đến n=k, ta có: \(Q=2^{2^k}+4^k+1⋮3\)
Với n=k+1 thì \(Q=2^{2^k.2}+4^{k+1}+1=2^{2^k}.2^{2^k}+4^k.4+1\)
\(=\left(2^{2^k}.2^{2^k}+2^{2^k}.4^k+2^{2^k}\right)-\left(2^{2^k}.4^k+2^{2^k}-4^k.4-4\right)-3\)
\(=2^{2^k}\left(2^{2^k}+4^k+1\right)-\left(4^k+1\right)\left(2^{2^k}-4\right)-3\)
\(=2^{2^k}Q-\left(4^k+1\right)\left(4^{2^{k-1}}-1-3\right)-3⋮3\) do \(\left(4^{2^{k-1}}-1\right)⋮\left(4-1\right)=3\)