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Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
a) \(\dfrac{2x-6}{x^2-x-6}\)
\(=\dfrac{2\left(x-3\right)}{x^2-3x+2x-6}\)
\(=\dfrac{2\left(x-3\right)}{x\left(x-3\right)+2\left(x-3\right)}\)
\(=\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\dfrac{2}{x+2}\)
b) \(\dfrac{6x^2-x-2}{4x^2-1}\)
\(=\dfrac{6x^2+3x-4x-2}{\left(2x\right)^2-1^2}\)
\(=\dfrac{3x\left(2x+1\right)-2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{\left(2x+1\right)\left(3x-2\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{3x-2}{2x-1}\)
\(c,\dfrac{x^3-x^2+3x-3}{x^3+2x^2+3x+6}\)
\(=\dfrac{x^2\left(x-1\right)+3\left(x-1\right)}{x^2\left(x+2\right)+3\left(x+2\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+3\right)}{\left(x+2\right)\left(x^2+3\right)}=\dfrac{x-1}{x+2}\)
d,Sửa đề :
\(\dfrac{a^2-b^2+c^2+2ac}{a^2+b^2-c^2+2ab}\)
\(=\dfrac{\left(a^2+2ac+c^2\right)-b^2}{\left(a^2+2ab+b^2\right)-c^2}\)
\(=\dfrac{\left(a+c\right)^2-b^2}{\left(a+b\right)^2-c^2}\)
\(=\dfrac{\left(a-b+c\right)\left(a+b+c\right)}{\left(a+b-c\right)\left(a+b+c\right)}\)
\(=\dfrac{a-b+c}{a+b-c}\)
e,g Đề ko rõ
a: \(\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+x^2+x+1\right)\)
\(=x^n+x^{n-1}+...+x^3+x^2+x-x^{n-1}-x^{n-2}-...-x^2-x-1\)
\(=x^n-1\)
b: \(\left(a-b\right)\left(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}\right)\)
\(=a^n+a^{n-1}b+...+a^2b^{n-2}+ab^{n-1}-a^{n-1}b-a^{n-2}b^2-...-ab^{n-1}-b^n\)
\(=a^n-b^n\)