\(400^n+256^n-9^n-1\)

Theo hằng đẳng thức 8 và 9

\(\Rightarrow400^n+256^n-9^n-1^n⋮\left(400+256-9-1\right)=646\)

Mà \(646⋮323\Rightarrow\left(đpcm\right)\)

I don't now

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

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

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

\(=3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)

d)  \(VT=a^3+b^3+c^3-3abc\)

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

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

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

I don't now

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\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow\)\(x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow\)\(2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\)\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\)\(\Leftrightarrow\)\(x=y=z\)

\(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

a) mk chỉnh đề:

Chứng minh:  \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)   (1)

                hoặc   \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\) (2)

            BÀI LÀM

TH1:

\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VP\)  (đpcm)

TH2:

\(VP=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2=VT\)  (đpcm)

b)  \(a+b=9\)\(\Rightarrow\)\(a=9-b\)

Ta có:  \(ab=20\)\(\Rightarrow\)\(\left(9-b\right).b=20\)

\(\Leftrightarrow\)\(b^2-9b+20=0\)

\(\Leftrightarrow\)\(\left(b-4\right)\left(b-5\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}b=4\\b=5\end{cases}}\)

Nếu  \(b=4\)thì:  \(a=5\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(5-4\right)^{2011}=1\)

Nếu  \(b=5\)thì  \(a=4\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(4-5\right)^{2011}=-1\)

a, sửa đề CM: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VT\left(đpcm\right)\)

b, \(a+b=9\Leftrightarrow\left(a+b\right)^2=81\Leftrightarrow\left(a-b\right)^2+4ab=81\Leftrightarrow\left(a-b\right)^2=81-4.20=1\Leftrightarrow a-b=\pm1\)

Với \(a-b=1\Rightarrow\left(a-b\right)^{2011}=1\)

Với \(a-b=-1\Rightarrow\left(a-b\right)^{2011}=-1\)

Ta có: \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{2a.2a.2a}{a.a.a}=\frac{8a^3}{a^3}=8\)

I don't now

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