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\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)=x^2\left(y-z\right)-y^2\left[\left(y-z\right)+\left(x-y\right)\right]+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(x^2-y^2\right)\left(y-z\right)-\left(y^2-z^2\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)-\left(y-z\right)\left(y+z\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x+y-y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
B = (x + 3)(x - 1)(x - 5)(x + 15) + 64x2
B = x4 + 12x3 - 58x2 - 180x + 225 + 64x2
B = x4 + 12x3 + 6x2 - 180x + 225
(x2 + x + 2)(x2 + 9x + 18) - 28
= x4 + 10x3 + 29x2 + 36x + 36 - 28
= x4 + 10x3 + 29x2 + 36x + 8
\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(x^4+6x^3+11x^2+6x+1\)
\(=\left(x^4+6x^3+9x^2\right)+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
Chúc bạn học tốt.
a, \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)\)
\(=\left[\left(x+2\right)\left(x-7\right)\right].\left[\left(x+3\right)\left(x-8\right)\right]\)
\(=\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144\)(1)
Đặt \(x^2-5x-14=t\) thì \(x^2-5x-24=t-10\)
Thay vào (1), ta có:
\(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)\)
\(=t\left(t-10\right)-144\)
\(=t^2-10t-144\)
\(=t^2-18t+8t-144\)
\(=t\left(t-18\right)+8\left(t-18\right)\)
\(=\left(t+8\right)\left(t-18\right)\)
\(=\left(x^2-5x-14+8\right)\left(x^2-5x-14-18\right)\)
\(=\left(x^2-5x-6\right)\left(x^2-5x-32\right)\)
\(=\left(x+1\right)\left(x-6\right)\left(x^2-5x-32\right)\)
a)\(\left(x-y\right)^2-2\left(x-y\right)+1=\left(x-y-1\right)^2\)
b)\(x^2-2y-1-2x+1-y^2=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left[\left(x-1\right)-\left(y+1\right)\right]\left[\left(x-1\right)+\left(y+1\right)\right]\)
\(=\left(x-y-2\right)\left(x+y\right)\)
c)\(x^2-y^2-2x-1=x^2-\left(y^2+2x+1\right)\)
\(=x^2-\left(y+1\right)^2\)
\(=\left(x^2-y-1\right)\left(x^2+y+1\right)\)
A. Ta có: (x - y)2 - 2(x - y)+1 = (x - y)2 - 2.(x - y).1 +12 = ( x - y - 1)2
B. Ta có: x2 - 2y -1 - 2x +1 -y2 = (x2 - y2) - (2x - 2y) -1+1 = (x - y)(x + y) - 2(x - y) = (x - y)(x + y - 2)
C. Ta có: x2 - y2 -2y -1 = x2 -(y2 - 2y -1) = x2 - ( y2 +2y1 + 1) = x2 - (y+1)2 = (x - y - 1)(x + y +1)
k cho mình nha bạn hihj!!! ~3~
a
Ta có
\(2x^2+2x=2x\left(x+1\right)\)
b
\(\left(1+xy\right)^2-\left(x+y\right)^2=\left(1+xy-x-y\right)\left(1+xy+x+y\right)\)
\(\left[\left(1-x\right)-y\left(1-x\right)\right]\left[\left(1+x\right)+y\left(1+x\right)\right]=\left(1-x\right)\left(1-y\right)\left(1+x\right)\left(1+y\right)\)
(x + y)3 - 1 - 3xy(x + y - 1)
= x3 + 3x2y + 3xy2 + y3 - 1 - 3x2y - 3xy2 + 3xy
= x3 - 1 + 3xy
= x(x2 + 3y) - 1
k bt lm nx r :v
\(\left(x+y\right)^3-1-3xy\left(x+y-1\right) \)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)