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Phân tích ra hai đa thức chứ cùng một đa thức là \(x^2+x+1\) nên P(x) chia hết cho Q(x) với x thuộc Z
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/ \(B=7.5^{2n}+12.6^n=\left(7.25^n-7.6^n\right)+19.6^n\)
\(=7\left(25-6\right)G\left(n\right)+19.6^n=7.19.G\left(n\right)+19.6^n⋮19\)
Lời giải:
Ta có:
\(A=x^{1970}+x^{1930}+x^{1980}=x^{1930}(x^{50}+x^{40}+1)\)
Xét \(x^{50}+x^{40}+1=x^{30}(x^{20}+x^{10}+1)-(x^{30}-1)\)
\(=x^{30}(x^{20}+x^{10}+1)-(x^{10}-1)(x^{20}+x^{10}+1)\)
\(=(x^{20}+x^{10}+1)(x^{30}-x^{10}+1)\vdots x^{20}+x^{10}+1\)
Vì \(x^{50}+x^{40}+1\vdots x^{20}+x^{10}+1\Rightarrow A\vdots x^{20}+x^{10}+1\)
Do đó ta có đpcm.
Phần a)
Sử dụng bổ đề \(x^{mn}-1\vdots x^m-1\) với mọi \(m,n \in\mathbb{N}\)
Chứng minh bổ đề:
Thật vậy, theo hằng đẳng thức đáng nhớ:
\(x^{mn}-1=(x^m)^n-1^n=(x^m-1)[(x^m)^{n-1}+(x^m)^{n-2}+...+x^m+1]\vdots x^m-1\)
Bổ đề đc chứng minh.
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Ta có:
\(x^{400}+x^{200}+1=x^{396}.x^4+x^{198}.x^2+1\)
\(=x^4(x^{396}-1)+x^2(x^{198}-1)+(x^4+x^2+1)\)
Áp dụng bổ đề trên vào bài toán kết hợp với \(x^6-1=(x^2-1)(x^4+x^2+1)\vdots x^4+x^2+1\) ta suy ra:
\(x^{396}-1=x^{6.66}-1\vdots x^6-1\vdots x^4+x^2+1\)
\(x^{198}-1=x^{6.33}-1\vdots x^6-1\vdots x^4+x^2+1\)
\(x^4+x^2+1\vdots x^4+x^2+1\) (hiển nhiên)
Do đó: \(x^{400}+x^{200}+1\vdots x^4+x^2+1\)
(đpcm)
Phần b)
\(F(x)=x^{1970}+x^{1930}+x^{1890}=x^{1890}(x^{80}+x^{40}+1)\)
Thấy rằng:
\(x^{80}+x^{40}+1=(x^{40}+1)^2-x^{40}=(x^{40}+1)^2-(x^{20})^2\)
\(=(x^{40}+1-x^{20})(x^{40}+1+x^{20})\)
Mà: \(x^{40}+1+x^{20}=(x^{20}+1)^2-x^{20}=(x^{20}+1)^2-(x^{10})^2\)
\(=(x^{20}+1-x^{10})(x^{20}+1+x^{10})\vdots x^{20}+x^{10}+1\)
Do đó:
\(x^{80}+x^{40}+1\vdots x^{20}+x^{10}+1\)
\(\frac{x+10}{2000}+\frac{x+20}{1990}+\frac{x+30}{1980}+\frac{x+40}{1970}=-4\)
\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)
\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)
Vì \(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}>0\)
\(\Rightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)
\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)
mà\(\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)\ne0\Rightarrow\left(x+2010\right)=0\\ \Rightarrow x=-2010\)
Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}-3=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}-3\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1+\frac{x-25}{1970}-1=\frac{x-1990}{5}-1+\frac{x-1980}{15}-1+\frac{x-1970}{25}-1\)\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow\)\(x-1995=0\)
\(\Leftrightarrow\)\(x=1995\)
x-5/1990+x-15/1980+x-25/1970=x-1990/5+x-1980/15+x-1970/25
<=> (x-5/1990-1)+(x-15/1980-1)+(x-25/1970-1)=(x-1990/5-1)+(x-1980/15-1)+(x-1970/25-1)
<=> x-1995/1990+x-1995/1980+x-1995/1970=x-1995/5+x-1995/15+x-1995/25
<=> (x-1995)(1/1990+1/1980+1/1970-1/5-1/15-1/25)=0
<=> x-1995=0
<=> x=1995