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a) \(\left(n+3\right)^2-\left(n-1\right)^2\)
\(=\left(n+3+n-1\right)\left(n+3-n+1\right)\)
\(=\left(2n+2\right)4\)
\(=2\left(n+1\right).4\)
\(=8\left(n+1\right)⋮8\)
=> đpcm
1) \(A=-2x^2-10y^2+4xy+4x+4y+2013=-2\left(x-y-1\right)^2-8\left(y-\frac{1}{2}\right)^2+2017\le2017\forall x,y\inℝ\)Đẳng thức xảy ra khi x = 3/2; y = 1/2
2) \(A=a^4-2a^3+2a^2-2a+2=\left(a^2+1\right)\left(a-1\right)^2+1\ge1\)
Đẳng thức xảy ra khi a = 1
3) \(N=\left(x-y\right)\left(x-2y\right)\left(x-3y\right)\left(x-4y\right)+y^4=\left(x^2-5xy+4y^2\right)\left(x^2-5x+6y^2\right)+y^4=\left(x^2-5xy+4y^2\right)^2+2y^2\left(x^2-5xy+4y^2\right)+y^4=\left(x^2-5xy+5y^2\right)^2\)(là số chính phương, đpcm)
4) \(a^3+b^3=3ab-1\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3ab+1=0\Leftrightarrow\left[\left(a+b\right)^3+1\right]-3ab\left(a+b+1\right)=0\)\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b+1\right)=0\Leftrightarrow\left(a+b+1\right)\left(a^2+b^2-ab-a-b+1\right)=0\)Vì a, b dương nên a + b + 1 > 0 suy ra \(a^2+b^2-ab-a-b+1=0\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\Leftrightarrow a=b=1\)
Do đó \(a^{2018}+b^{2019}=1+1=2\)
5) \(A=n^3+\left(n+1\right)^3+\left(n+2\right)^3=3n\left(n^2+5\right)+9\left(n^2+1\right)⋮9\)(Do số chính phương chia 3 dư 1 hoặc 0)
Lan nghĩ ra một số biết rằng số đó bằng hiệu của số chẵn lớn nhất có 3 chữ số chẵn khác nhau với 60 rồi cộng thêm 21. Hỏi số lan nghĩ là số nào
1/
\(\dfrac{\left(x-y\right)^3-3xy\left(x+y\right)+y^3}{x-6y}\)
\(=\dfrac{x^3-3x^2y+3xy^2-y^3-3x^2y-3xy^2+y^3}{x-6y}\)
\(=\dfrac{x^3-6x^2y}{x-6y}\)
\(=\dfrac{x^2\left(x-6y\right)}{x-6y}\)
\(=x^2\)
\(2\)/
\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\dfrac{\left(x-y+z^{ }\right)^2}{\left(x-y\right)^2-z^2}\)
\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}\)
\(=\dfrac{x-y+z}{x-y-z}\)
3/
\(\dfrac{\left(n+1\right)!}{n!\left(n+2\right)}\)
\(=\dfrac{n!\left(n+1\right)}{n!\left(n+2\right)}\)
\(=\dfrac{n+1}{n+2}\)
4/
\(\dfrac{n!}{\left(n+1\right)!-n!}\)
\(=\dfrac{n!}{n!\left(n+1\right)-n!}\)
\(=\dfrac{n!}{n!\left[\left(n+1\right)-1\right]}\)
\(=\dfrac{n!}{n!.n}\)
\(=\dfrac{1}{n}\)
5/
\(\dfrac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}\)
\(=\dfrac{\left(n+1\right)!-\left(n+1\right)!\left(n+2\right)}{\left(n+1\right)!+\left(n+1\right)!\left(n+2\right)}\)
\(=\dfrac{\left(n+1\right)!\left(-n-1\right)}{\left(n+1\right)!\left(n+3\right)}\)
\(=\dfrac{-n-1}{n+3}\)
Ta có :
\(\frac{n\left(n+1\right)}{2}+\frac{\left(n+1\right)\left(n+2\right)}{2}\)
\(=\frac{n\left(n+1\right)+\left(n+1\right)\left(n+2\right)}{2}\)
\(=\frac{\left(n+1\right)\left(n+n+2\right)}{2}\)
\(=\frac{\left(n+1\right)\cdot2\cdot\left(n+1\right)}{2}\)
\(=\left(n+1\right)^2\)
=> ĐPCM
vì bài dài quá nên mình làm từng bài 1 nhé
1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Do đó :
\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)
2.
Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Do đó :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)
toi ko bt