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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) \(2xy^3-6x^2+10xy\)
\(=2x.y^3-2x.3x+2x.5y\)
\(=2x\left(y^3-3x+5y\right)\)
\(=2x[y\left(y^2-5\right)-3x]\)
bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)
\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)
Bài 2:
1: \(x^2y^2-8-1\)
\(=x^2y^2-9\)
\(=\left(xy-3\right)\left(xy+3\right)\)
2: \(x^3y-2x^2y+xy-xy^3\)
\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)
\(=xy\left(x^2-2x+1-y^2\right)\)
\(=xy\left[\left(x-1\right)^2-y^2\right]\)
\(=xy\left(x-1-y\right)\left(x-1+y\right)\)
3: \(x^3-2x^2y+xy^2\)
\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)
\(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
4: \(x^2+2x-y^2+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
5: \(x^2+2x-4y^2+1\)
\(=\left(x^2+2x+1\right)-4y^2\)
\(=\left(x+1\right)^2-4y^2\)
\(=\left(x+1-2y\right)\left(x+1+2y\right)\)
6: \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
(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)
1) \(x^2-4xy+4y^2+xz-2yz\)
\(=\left(x^2-4xy+4y^2\right)+\left(xz-2yz\right)\)
\(=\left(x-2y\right)^2+z\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-2y+z\right)\)
2) \(\left(x-y\right)^3+\left(x+y\right)^3\)
\(=\left[\left(x-y\right)+\left(x+y\right)\right]\left[\left(x-y\right)^2-\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\right]\)
\(=\left(x-y+x+y\right)\left(x^2-2xy+y^2-x^2+y^2+x^2+2xy+y^2\right)\)
\(=2x\left(x^2+3y^2\right)\)
1: Đa thức này ko phân tích được nha bạn
2: \(x^2+8x+7\)
\(=x^2+x+7x+7\)
\(=x\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(x+7\right)\)
3: \(x^2-6x-16\)
\(=x^2-8x+2x-16\)
\(=x\left(x-8\right)+2\left(x-8\right)\)
\(=\left(x-8\right)\left(x+2\right)\)
4: \(4x^2-8x+3\)
\(=4x^2-2x-6x+3\)
\(=2x\left(2x-1\right)-3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x-3\right)\)
5: \(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
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