Thành Nhân Tử
Bài 1:
a, \(9x^2+6x-8\)
\(=9x^2+12x-6x-8\)
\(=3x\left(3x+4\right)-2\left(3x+4\right)=\left(3x-2\right)\left(3x+4\right)\)
b, \(x^2-7xy+10y^2\)
\(=x^2-5xy-2xy+10y^2\)
\(=x\left(x-5y\right)-2y\left(x-5y\right)=\left(x-2y\right)\left(x-5y\right)\)
c, \(a^4+4b^4\)
\(=a^4+4a^2b^2+4b^4-4a^2b^2\)
\(=\left(a^2+2b^2\right)^2-4a^2b^2\)
\(=\left(a^2-2ab+2b^2\right)\left(a^2+2ab+2b^2\right)\)
d, \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
\(=x^4+5x^3+6x^2+5x^3+25x^2+30x+4x^2+20x+24-24\)
\(=x^4+10x^3+35x^2+50x\)
\(=x^4+5x^3+5x^3+25x^2+10x^2+50x\)
\(=x^3\left(x+5\right)+5x^2\left(x+5\right)+10x\left(x+5\right)\)
\(=\left(x+5x^2+10x\right)\left(x+5\right)\)
e, \(x^2+19x+84\)
\(=x^2+7x+12x+84\)
\(=x\left(x+7\right)+12\left(x+7\right)=\left(x+12\right)\left(x+7\right)\)
Bài 2:
a, \(x^2-25+3\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+3\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-15+x+5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\4x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
Vậy...
b, \(x^3-3x^2+3x-1=1\)
\(\Leftrightarrow x^3-2x^2-x^2+2x+x-2=0\)
\(\Leftrightarrow x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x-2\right)=0\)
Dễ thấy \(x^2-x+1\ne0\forall x\)
\(\Rightarrow x-2=0\Leftrightarrow x=2\)
Vậy x = 2
Bài 3:
a, \(A=x^2+x+5=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{19}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu " = " khi \(\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{-1}{2}\)
Vậy \(MIN_A=\dfrac{19}{4}\) khi \(x=\dfrac{-1}{2}\)
b, tương tự
