Phân tích đa thức thanh nhân tử
a, 3x33 + 9x
b, xy + y2 - 3x - 3y
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\(a,\left(x+2\right)-\left(x-2\right).\left(x+2\right)\)
\(=\left(x+2\right).\left(1-x+2\right)=\left(x+2\right).\left(-x+3\right)\)
\(=-x^2+x+6\)
\(b,3x^2-6x=0\)
\(3x.\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(-3x^2-7x+10\)
\(=3x-3x^2+10-10x\)
\(=3x.\left(1-x\right)+10.\left(1-x\right)=\left(3x+10\right).\left(1-x\right)\)
\(-3x^2+3y^2-4xz-4yz\)
\(=3\left(y^2-x^2\right)-4z\left(x+y\right)\)
\(=3\left(y-x\right)\left(x+y\right)-4z\left(x+y\right)\)
\(=\left(x+y\right)\left(3y-3x-4z\right)\)
\(B1,a,A=x^2-6x+11\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\ge2\)
Dấu "=" <=> x=3
Vậy ..........
\(b,B=x^2-20x+101\)
\(=\left(x^2-20x+100\right)+1\)
\(=\left(x-10\right)^2+1\ge1\)
Dấu "=" <=> x = 10
Vậy .
\(2,a,A=4x-x^2+3\)
\(=7-\left(x^2-4x+4\right)\)'
\(=7-\left(x-2\right)^2\le7\)
Dấu ''='' <=> x = 2
Vậy .
\(b,B=-x^2+6x-11\)
\(=-2-\left(x^2-6x+9\right)\)
\(=-2-\left(x-3\right)^2\le-2\)
Dấu ""=" <=> x = 3
Vậy..
\(\left(3x+12\right)-\left(x^2-16\right)=3\left(x+4\right)-\left(x+4\right)\left(x-4\right)=\left(x+4\right)\left(3-x+4\right)\)
=(x+4)(7-x)
a)\(3.3x^3+9x\)
\(=9x.\left(x^2+1\right)\)
b) \(xy+y^2-3x-3y\)
\(y.\left(x+y\right)-3.\left(x+y\right)=\left(x+y\right).\left(y-3\right)\)
\(3x^33+9x=9x\left(x^2+1\right)\)
\(xy+y^2-3x-3y=y\left(x+y\right)-3\left(x+y\right)=\left(y-3\right)\left(x+y\right)\)