\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=[\left(a+b\right)^3+c^3]-[3ab\left(a+b\right)+3abc]=\left(a+b+c\right)[\left(a+b\right)^2-\left(a+b\right)c+c^3]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-3ab-ab-bc-ca\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

tick minh nha ban

thank you ban nhieu

\(A=x^3-3xy-y^3=x^3-3xy\left(x-y\right)-y^3\) ( vi x - y = 1)

\(=x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3=1^3=1\)

Cx như a, \(x=\dfrac{1-y}{3}\) thay vào N đc:

\(N=xy=\dfrac{1-y}{3}.y=\dfrac{y-y^2}{3}=\dfrac{-\left(y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}\right)}{3}\)

\(=\dfrac{-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{3}\)

Bn tự chứng minh \(-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\). Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(y-\dfrac{1}{2}\right)=0\Leftrightarrow y=\dfrac{1}{2}\)

\(\Rightarrow N=\dfrac{-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{3}\le\dfrac{\dfrac{-3}{4}}{3}=\dfrac{-1}{4}\)

Vậy MAX N = \(\dfrac{-1}{4}\Leftrightarrow y=\dfrac{1}{2}\)

a,Từ \(3x+y=1\Rightarrow x=\dfrac{1-y}{3}\)

\(\Rightarrow M=3x^2+y^2=3.\left(\dfrac{1-y}{3}\right)^2+y^2=3.\dfrac{y^2-2y+1}{9}+y^2\)

\(=\dfrac{3y^2+y^2-2y+1}{3}=\dfrac{4y^2-2y+1}{3}\)

Ta có: \(4y^2-2y+1=4y^2-2.2y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Bn tự chứng minh \(\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2y-\dfrac{1}{2}\right)^2=0\Leftrightarrow y=\dfrac{1}{4}\)

\(\Rightarrow M=\dfrac{\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{3}\ge\dfrac{\dfrac{3}{4}}{3}=\dfrac{1}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\) \(y=\dfrac{1}{4}\);\(x=\dfrac{1-\dfrac{1}{4}}{3}=\dfrac{1}{4}\)

Ta có: \(A=n^2+n+9=n^2+4n-3n-12+21=\left(n+4\right)\left(n-3\right)+21\)

Do \(\left(n+4\right)-\left(n-3\right)=7⋮7\) nên \(n+4\) và \(n-3\) có cùng số dư khi chia cho 7.Vậy có hai trường hợp: hoặc \(n+4\) và \(n-3\) cùng chia hết cho 7 hoặc n + 4 và n - 3 ko cùng chia hết cho 7

TH1: Suy ra \(\left(n+4\right)\left(n-3\right)⋮7\) mà 21 \(⋮̸\)49 \(\Rightarrowđpcm\) (1)

TH2: Suy ra \(\left(n+4\right)\left(n-3\right)\) ko chia hết cho 7 nhưng 21 chia hết cho 7 suy ra A ko chia hết cho 7 suy ra A ko chia hết cho 49 (2)

(1);(2)\(\Rightarrowđpcm\\\)

\(A=x+x^3+x^{27}+x^{2017}=\left(x^{2017}+x^3\right)+\left(x^{27}+x\right)\)

\(=x^3\left(x^{2014}+1\right)+x\left(x^{26}+1\right)=x^3\left(\left(x^2\right)^{1007}+1\right)+x\left(\left(x^2\right)^{13}+1\right)\)Ta có \(x^3\left(\left(x^2\right)^{1007}+1\right)⋮x^2+1\) và \(x\left(\left(x^2\right)^{13}+1\right)⋮x^2+1\)

\(\Rightarrow A=x^3\left(\left(x^2\right)^{1007}+1\right)+x\left(\left(x^2\right)^{13}+1\right)⋮x^2+1\)

Do đó số dư khi chia A cho \(x^2+1\) là 0

\(4x^2+4xy+y^2=\left(2x+y\right)^2=\left(2.2-6\right)^2=\left(-2\right)^2=4\)

\(A=31^n-15^n-24^n+8^n=\left(31^n-15^n\right)-\left(24^n-8^n\right)\)

\(=BS16-BS16=BS16⋮16\) (1)

\(A=\left(31^n-24^n\right)-\left(15^n-8^n\right)=BS7-BS7=BS7⋮7\) (2)

Mà (16,7) = 1; 112 = 16.7 \(\Rightarrow A⋮112\)