phân tích đa thức
x2-6x+10
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a) \(\dfrac{x+6}{x^2-4}+\dfrac{1}{x+2}=\dfrac{x+6}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+6+x-2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{2x+4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b) \(x^2+xy-5\left(x+y\right)=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\)
1.
\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)
2.
\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt
Bài 1:
Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=6x\left(2x+1\right)\cdot2y\)
\(=12xy\left(2x+1\right)\)
Bài 2:
Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
\(3x^4+6x^3-7x^2+8x-10\)
\(=\left(3x^4-3x^3\right)+\left(9x^3-9x^2\right)+\left(2x^2-2x\right)+\left(10x-10\right)\)
\(=\left(x-1\right)\left(3x^3+9x^2+2x+10\right)\)
a) \(x^3+6x^2+3x-10\)
\(=x^3-x^2+7x^2-7x+10x-10\)
\(=x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+7x+10\right)\)
\(=\left(x-1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x+5\right)\)
b) \(x^3+3x^2-33x-35\)
\(=x^3-5x^2+8x^2-40x+7x-35\)
\(=x^2\left(x-5\right)+8x\left(x-5\right)+7\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+8x+7\right)\)
\(=\left(x-5\right)\left(x^2+x+7x+7\right)\)
\(=\left(x-5\right)\left(x+1\right)\left(x+7\right)\)
Ta có: \(1+6x-6x^2-x^3\)
\(=-x^3-6x^2+6x+1\)
\(=\left(-x^3+1\right)-6x\left(x-1\right)\)
\(=-\left(x-1\right)\left(x^2+x+1\right)-6x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\left(-x^2-x-1-6x\right)\)
\(=-\left(x-1\right)\left(x^2+7x+1\right)\)
đặt y=x2+1
=>y2=(x2+1)2
y2=x4+2x2+1
đặt P(x)=x^4+6x^3+11x^2+6x+1
=x4+2x2+1+6x3+6x+9x2
=x4+2x+1+6x(x2+1)+9x2
thay y2=x4+2x2+1 và y=x2+1 ta được
Q(y)=y2+6xy+9x2
=(y+3x)2
thay y=x2+1 ta được:
(x2+3x+1)2
vậy x^4+6x^3+11x^2+6x+1=(x2+3x+1)2
Đề sai hả bạn??