Câu 2c bn ghi kết luận giùm mik là n€ {0;1}

Đáp án:a)n=0;1

8a)x+2

b)\(\frac{2a}{a^{2}+ab+b^{2}}\)

Giải thích các bước giải:

7c)\( n^{5}+1\vdots n^{3}+1⇒ n^{6}+n\vdots n^{3}+1\)

ta có: \(\frac{n^{6}+n}{n^{3}+1}=\frac{n^{6}-1+n+1}{n^{3}+1}=\frac{n^{6}-1}{n^{3}+1}+\frac{n+1}{n^{3}+1}\)
=\(\frac{(n^{3})^{2}-1}{n^{3}+1}+\frac{n+1}{(n+1)(n^{2}-n+1)}\)
=\(\frac{(n^{3}-1)(n^{3}+1)}{n^{3}+1}+\frac{1}{n^{2}-n+1}\)
=\(n^{3}+1+\frac{1}{n^{2}-n+1}\)

Đer pt trên chia hết thì 1 chia hết \(n^{2}-n+1\)

⇒\( n^{2}-n+1=1⇒ n=0;1\)

8a) \((2x^{3}+4x^{2}+5x+10):(2x^{2}+5)=[(2x^{3}+4x^{2})+(5x+10)]:(2x^{2}+5)\)
=\([2x^{2}(x+2)+5(x+2)]:(2x^{2}+5)=(2x^{2}+5)(x+2):(2x^{2}+5)=x+2\)

b)\(\frac{a+b}{a^{2}+b^{2}+ab}+\frac{1}{a-b}-\frac{3ab}{a^{3}-b^{3}}\)
=\(\frac{a+b}{a^{2}+b^{2}+ab}+\frac{1}{a-b}-\frac{3ab}{(a-b)(a^{2}+ab+b^{3})}\)
=\(\frac{(a+b)(a-b)+a^{2}+ab+b^{2}-3ab}{(a-b)(a^{2}+ab+b^{3})}\)
=\(\frac{a^{2}-b^{2}+a^{2}+ab+b^{2}-3ab}{(a-b)(a^{2}+ab+b^{3})}\)
=\(\frac{2a^{2}-2ab}{(a-b)(a^{2}+ab+b^{3})}=\frac{2a(a-b)}{(a-b)(a^{2}+ab+b^{3})}\)
=\(\frac{2a}{a^{2}+ab+b^{2}}\)