phân tích thành nhân tử
6x4+5x3-38x2+5x+6
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f ) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(x^2+5x+5=t\), ta có :
\(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
Thay và ta có :
\(\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
a) \(24x^2-4xy\)
\(=4x\left(6x-y\right)\)
b) \(5x^3-10x^2+5x-20xy^2\)
\(=5x\left(x^2-10x+5-20y^2\right)\)
\(a,=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ b,=\left(x+y\right)\left(x-5\right)\\ c,=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2y\right)\left(x-y\right)\\ d,=x^2-2xy=x\left(x-2y\right)\\ e,=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
\(=5x\left(x^2-2xy+y^2\right)\)
\(=5x\left(x-y\right)^2\)
a) \(5x^3-10x^2+15x=5x\left(x^2-2x+3\right)\)
b) \(x^2-3x+2=x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(x-1\right)\)
\(6x^4+5x^3-38x^2+5x+6\)
\(=6x^4-15x^3+6x^2+20x^3-50x^2+20x+6x^2-15x+6\)
\(=3x^2\left(2x^2-5x+2\right)+10x\left(2x^2-5x+2\right)+3\left(2x^2-5x+2\right)\)
\(=\left(2x^2-5x+2\right)\left(3x^2+10x+3\right)\)
\(=\left(2x^2-x-4x+2\right)\left[3x^2+x+9x+3\right]\)
\(=\left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\)
\(=\left(x-2\right)\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\)