phân tích đa thức thành nhân tử
1/ (x -1)(x - 2)(x + 4)(x + 5) - 112
2 / (x -2)(x + 2)( x^2 - 10 ) - 72
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\(3x^2+x-2=3x^2-2x+3x-2=x\left(3x-2\right)+\left(3x-2\right)=\left(x+1\right)\left(3x-2\right)\)
\(x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(x^2+2xy-15y^2=x^2-3xy+5xy-15y^2=x\left(x-3y\right)+5y\left(x-3y\right)=\left(x+5y\right)\left(x-3y\right)\)
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
1) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
3) \(x^2+7x+12=\left(x^2+3x\right)+\left(4x+12\right)=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
1) \(2xy^3-6x^2+10xy\)
\(=2x.y^3-2x.3x+2x.5y\)
\(=2x\left(y^3-3x+5y\right)\)
\(=2x[y\left(y^2-5\right)-3x]\)
(x-1)(x-2)(x+4)(x+5)-72=[(x-1)(x+4)][x-2)(x+5)]-72=(x^2+3x-4)(x^2+3x-10)-72
Đặt x^2+3x-4=t nên x^2+3x-10=t-6. Thay vào (*) ta được :
(x-1)(x-2)(x+4)(x+5)=t.(t-6)-72=t^2-6t-72=t^2-6t+9-81=(t-3)^2-9^2=(t-3-9)(t-3+9)=(t-12)(t+6)=(x^2+3x-16)(x^2+3x+2)
Câu 1:
\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)-112\)
\(=\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+5\right)-112\)
\(=\left(x^2+3x-4\right)\left(x^2+3x-10\right)-112\)
\(=\left(x^2+3x-7\right)^2-3^2-112\)
\(=\left(x^2+3x-7\right)^2-11^2\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-18\right)\)
\(=\left(x^2+3x+4\right)\left(x+6\right)\left(x-3\right)\)
Câu 2:
\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)-72\)
\(=\left(x^2-4\right)\left(x^2-10\right)-2\)
\(=\left(x^2-7\right)^2-3^2-72\)
\(=\left(x^2-7\right)^2-81\)
\(=\left(x^2-16\right)\left(x^2+2\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
(x−1)(x−2)(x+4)(x+5)−112
=(x−1)(x+4)(x−2)(x+5)−112
=(x^2+3x−4)(x^2+3x−10)−112
=(x^2+3x−7)^2−32−112
=(x^2+3x−7)^2−112
=(x^2+3x+4)(x^2+3x−18)
=(x^2+3x+4)(x+6)(x−3)
Câu 2:
(x−2)(x+2)(x^2−10)−72
=(x2−4)(x^2−10)−2
=(x^2−7)^2−32−72