n = 1 ta thấy thảo mãn

Nếu \(n\ge2\)thì \(n^{1988}+n^{1987}+1>n^2+n+1\)

Mặt khác \(n^{1988}+n^{1987}+1=n^2\left(n^{1986}-1\right)+n\left(n^{1986}-1\right)+\left(n^2+n+1\right)\)

Nên \(n^2+n+1\)|\(n^{1988}+n^{1987}+1\)

Vậy \(n^{1988}+n^{1987}+1\)là hợp số

thoả mãn ko phải thảo mãn

Các bạn giải nhanh giúp mình nhé !

a) x=y=0

b) n bằng 0

câu a    x,y cùng bằng 0

câu b    n thuộc rỗng

n=1 nha bạn k cho mình nha

ta có : \(A=n^{1988}+n^{1987}+1\)

\(\Rightarrow A=n^2\left[\left(n^{662}\right)^3-1\right]+n\left[\left(n^{662}\right)^3-1\right]+\left(n^2+n+1\right)\)

mà \(\left(n^{662}\right)^3-1⋮\left(n^3-1\right)\)và \(n^3-1=\left(n-1\right)\left(n^2+n+1\right)\Rightarrow n^3-1⋮\left(n^2+n+1\right)\)

nên \(\left(n^{662}\right)^3-1⋮\left(n^2+n+1\right)\)

\(\Rightarrow A⋮n^2+n+1\)

Mặt khác : A là số nguyên tố 

=>\(\orbr{\begin{cases}n^2+n+1=1\\n^2+n+1=n^{1988}+n^{1987}+1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}n\left(n+1\right)=0\\n^2+n=n^{1986}\left(n^2+n\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}n=0;n=-1\\n\left(n+1\right)\left(n^{1986}-1\right)=0\end{cases}}\)

=> \(n\left(n+1\right)\left(n^{1986}-1\right)=0\) vì n nguyên dương

\(\Rightarrow n^{1986}-1=0\Rightarrow n=1\) (thỏa mãn)

thử lại : thay n=1 vào A ta đc : A= 1+1+1=3 là số nguyên tố

Vậy n=1 thì A là số nguyên tố

∙∙ n=1n=1 ta thấy thõa mãn

Nếu n≥2n≥2 thì n1998+n1987+1>n2+n+1n1998+n1987+1>n2+n+1

Mặt khác n1988+n1987+1=n2(n1986−1)+n(n1986−1)+(n2+n+1)n1988+n1987+1=n2(n1986−1)+n(n1986−1)+(n2+n+1)

Nên n2+n+1|n1988+n1987+1n2+n+1|n1988+n1987+1

Vậy n1988+n1987+1n1988+n1987+1 là hợp số

ủng hộ nhá

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n≥2" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n≥2 thì n1998+n1987+1>n2+n+1

n1988+n1987+1=n2(n1986−1)+n(n1986−1)+(n2+n+1)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n1988+n1987+1=n2(n1986−1)+n(n1986−1)+(n2+n+1)

n2+n+1|n1988+n1987+1" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n2+n+1|n1988+n1987+1

n1988+n1987+1" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n1988+n1987+1 là hợp số

Nếu \(n\ge2\)thì \(n^{1998}+n^{1987}+1>n^2+n+1\)

Mặt khác : \(n^{1998}+n^{1987}+1=n^2\left(n^{1986}-1\right)+n\left(n^{1986}-1\right)+\left(n^2+n+1\right)\)

Nên : \(n^2+n+1\)\(n^{1988}+n^{1987}+1\)

Vậy : \(n^{1998}+n^{1987}+1\)là hợp số 

 ta thấy thõa mãn

+) $n\ge 2$ thì $n^{1998}+n^{1987}+1>n^2+n+1$

Mặt khác $n^{1988}+n^{1987}+1=n^2(n^{1986}-1)+n(n^{1986}-1)+(n^2+n+1)$

Nên $n^2+n+1|n^{1988}+n^{1987}+1$

Vậy $n^{1988}+n^{1987}+1$ là hợp số

+) Với \(n=1\Rightarrow B=3\) là SNT

+) Với \(n>1\Rightarrow B>3\)

Ta có: \(B=\left(n^{1988}-n^2\right)+\left(n^{1987}-n\right)+\left(n^2+n+1\right)\)

Có \(n^{1986}-1=\left[\left(n^3\right)^{662}-1\right]⋮n^3-1\)

Mà \(n^3-1=\left(n-1\right)\left(n^2+n+1\right)\)

\(\Rightarrow n^{1986}-1⋮n^2+n+1\)

Mà \(\left\{{}\begin{matrix}n^{1988}-n^2⋮n^{1986}-1\\n^{1887}-n⋮n^{1986}-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n^{1988}-n^2⋮n^2+n+1\\n^{1987}-n⋮n^2+n+1\end{matrix}\right.\)

\(\Rightarrow B⋮n^2+n+1\)

Mà \(n^2+n+1>3\forall n>1\)

=> B ko là SNT với n > 1

Vậy n = 1 (T/m)

ta thấy thõa mãn

+) n≥2 thì n1998+n1987+1>n2+n+1

Mặt khác n1988+n1987+1=n2(n1986−1)+n(n1986−1)+(n2+n+1)

Nên n2+n+1|n1988+n1987+1

Vậy n1988+n1987+1 là hợp số