Phân tích đa thức thành nhân tử sử dụng phương pháp tách hạng tử thành nhiều hạng tử:
a) x3 + 3x2 + 8x + 4
b) 4x3 - 13x2 +9x -18
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\(400^n+256^n-9^n-1\)
Theo hằng đẳng thức 8 và 9
\(\Rightarrow400^n+256^n-9^n-1^n⋮\left(400+256-9-1\right)=646\)
Mà \(646⋮323\Rightarrow\left(đpcm\right)\)
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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