bài 1:
a,\(x^2+6x+5=x^2+5x+x+5=x\left(x+5\right)+\left(x+5\right)=\left(x+5\right)\left(x+1\right)\)b,\(x^4+2007x^2+2006x+2007=x^4-x+2007x^2+2007x+2007=x\left(x^3-1\right)+2007\left(x^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2007\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2007\right)\)
c,\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
đặt \(x^2+5x+5=y\) ta đc:
\(\left(y-1\right)\left(y+1\right)+1=y^2-1+1=y^2=\left(x^2+5x+5\right)^2\)
Câu 1:
\(\text{a) }x^2+6x+5\\ \\=x^2+5x+x+5\\ \\ =\left(x^2+5x\right)+\left(x+5\right)\\ \\ =x\left(x+5\right)+\left(x+5\right)\\ \\ =\left(x+1\right)\left(x+5\right)\)
\(\text{b) }x^4+2007x^2+2006x+2007\\ \\ =x^4+2007x^2+2007x-x+2007+x^3-x^3+x^2-x^2\\ \\ =\left(x^4-x^3+2007x^2\right)+\left(x^3-x^2+2007x\right)+\left(x^2-x+2007\right)\\ =x^2\left(x^2-x+2007\right)+x\left(x^2-x+2007\right)+\left(x^2-x+2007\right)\\ =\left(x^2+x+1\right)\left(x^2-x+2007\right)\)
\(\text{c) }\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\\ \\ =\left(x^2+x+4x+4\right)\left(x^2+2x+3x+6\right)+1\\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\left(1\right)\)
Đặt \(t=x^2+5x+5\left(\text{*}\right)\)
Thay \(\left(\text{*}\right)\) vào \(\left(1\right)\), ta được: \(\left(1\right)=\left(t-1\right)\left(t+1\right)+1\)
\(=t^2-1+1\\ \\ =t^2\left(2\right)\)
Thay \(\left(\text{*}\right)\) vào \(\left(1\right)\), ta được: \(\left(2\right)=\left(x^2+5x+5\right)^2\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+ab^2+ac^2-a^2b-a^2c-abc+a^2b+b^3+bc^2-ab^2-abc-b^2c+a^2c+a^2b+c^3-abc-ac^2-bc^2=0\)
\(\Leftrightarrow a\left(a^2+b^2+c^2-ab-ac-bc\right)+b\left(a^2+b^2+c^2-ab-ac-bc\right)+c\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\left(loai\right)\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{2}\left(a^2-2ab+b^2\right)+\dfrac{1}{2}\left(b^2-2bc+c^2\right)+\dfrac{1}{2}\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
Vậy t/g ABC là t/g đều
Câu 2:
\(\text{Ta có : }a^3+b^3+c^3=3abc\\ \Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Vì \(a;b;c\) là \(3\) cạnh của \(1\Delta\)
nên \(a+b+c\ne0\)
\(\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\\ \Rightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\ \Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0\forall a;b\)
\(\left(b-c\right)^2\ge0\forall b;c\\ \left(a-c\right)^2\ge0\forall a;c\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a;b;c\)
Dấu "=" xảy ra khi \(:\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Rightarrow a=b=c\)
Vậy \(\Delta ABC\) là \(\Delta\) đều
