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a) \(25.\left(x-1\right)^2-16\left(x+y\right)^2\)
= \(\left(5x-5\right)^2-\left(4x+y\right)^2\)
= \(\left(5x-5-4x-y\right)\left(5x-5+4x+y\right)\)
= \(\left(x-y-5\right)\left(9x+y-5\right)\)
b) \(x^3+3x^2+3x+1-27z^3\)
= \(\left(x+1\right)^3-27z^3\)
= \(\left(x+1-3z\right)\left(x^2+x.3z+9z^2\right)\)
c) \(x^2-2xy+y^2-xz+yz\)
= \(\left(x-y\right)^2-z\left(x-y\right)\)
= \(\left(x-y\right)\left(x-y-z\right)\)
d) \(a^3x-ab+b-x\)
= \(x\left(a^3-1\right)-b\left(a-1\right)\)
= \(x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)
= \(\left(a-1\right)\left(a^2x+ax+x-b\right)\)
f) \(x^2+2x-4y^2-4y\)
= \(x^2+2x+1-\left(4y^2+4y+1\right)\)
= \(\left(x+1\right)^2-\left(2y+1\right)^2\)
= \(\left(x+1-2y-1\right)\left(x+1+2y+1\right)\)
= \(\left(x-2y\right)\left(x+2y+2\right)\)
g) \(xy-4+2x-2y\)
= \(y\left(x-2\right)-2\left(x-2\right)\)
= \(\left(x-2\right)\left(y-2\right)\)
a: \(=\left(5x-5\right)^2-\left(4x-4y\right)^2\)
\(=\left(5x-5-4x+4y\right)\cdot\left(5x-5+4x-4y\right)\)
\(=\left(x+4y-5\right)\left(9x-4y-5\right)\)
b: \(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
c: \(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
d: \(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=x\left(a-1\right)\cdot\left(a^2+a+1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2x+ax+1-b\right)\)
k) \(x^3-x+3x^2+3xt^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
h) \(a^3-a^2x-ay+xy\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a^2-y\right)\left(a-x\right)\)
Câu c) Sử dụng hằng đẳng thức+Đặt biến phụ
Ta có: \(x^2+2xy+y^2-x-y-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y\right)\left(x+y-1\right)-12\)
Đặt: \(x+y=t\)
\(=t\left(t-1\right)-12\)
\(=t^2-t-12\)
\(=t^2-t-9-3\)
\(=\left(t^2-3^2\right)-\left(t+3\right)\)
\(=\left(t+3\right)\left(t-3\right)-\left(t+3\right)\)
\(=\left(t+3\right)\left(t-4\right)\)Bn tự thế vào nhá. (Bài c) tương tự bài a))
Câu d) Đặt biến phụ
Ta có: \(\left(5x^2-2x\right)^2+2x-5x^2-6\)
\(=\left(5x^2-2x\right)^2-5x^2+2x-6\)
\(=\left(5x^2-2x\right)^2-\left(5x^2-2x\right)-6\)
\(=\left(5x^2-2x\right)\left(5x^2-2x-1\right)-6\)
Đặt \(t=5x^2-2x\)
\(=t\left(t-1\right)-6\)
\(=t^2-t-6\)
\(=t^2-t-9+3\)
\(=\left(t^2-3^2\right)-\left(t-3\right)\)
\(=\left(t-3\right)\left(t+3\right)-\left(t-3\right)\)
\(=\left(t-3\right)\left(t+2\right)\)Bn tự thế t vào
Câu a) Sử dụng phương pháp đặt biến phụ+hằng đẳng thức
Ta có: \(\left(2x^2+x-2\right)\left(2x^2+x-3\right)-12\)
Đặt: \(t=2x^2+x-2\)
\(=t\left(t-1\right)-12\)
\(=t^2-t-12=t^2-t-9-3\)
\(=\left(t^2-3^2\right)-\left(t+3\right)\)
\(\left(t+3\right)\left(t-3\right)-\left(t+3\right)=\left(t+3\right)\left(t-4\right)\)
Thay t vào: \(\left(2x^2+x+1\right)\left(2x^2+x-6\right)\)
Câu b) Sử dụng hằng đẳng thức+ đặt biến phụ
Ta có: \(x^2+9y^2-9y-3x+6xy+2\)
\(=\left(x^2+6xy+9y^2\right)-\left(9y+3x\right)+2\)
\(=\left(x+3y\right)^2-3\left(3y+x\right)+2\)
\(=\left(x+3y\right)\left(x+3y-3\right)+2\)
Đặt \(t=x+3y\)
\(=t\left(t-3\right)+2\)
\(=t^2-3t+2\)
\(=\left(t^2-4\right)-\left(3t-6\right)\)
\(=\left(t-2\right)\left(t+2\right)-3\left(t-2\right)\)
\(=\left(t-2\right)\left(t-1\right)\)Khúc sau bn tự thế vào
Còn mấy bài sau đang nghiên cứu
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
\(e,-5x+x^2-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x+2\right)\left(x-7\right)\)
\(f,x^3+8+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+8x+4\right)\)
\(g,15x^2-7xy-2y^2\)
\(=15x^2+3xy-10xy-2y^2\)
\(=3\left(5x+y\right)-2y\left(5x+y\right)\)
\(=\left(5x+y\right)\left(3-2y\right)\)
\(h,3x^2-16x+5\)
\(=3x^2-x-15x+5\)
\(=x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x+5\right)\)
\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)
\(=x\left(x+y\right)^2\)
\(b,4x^2-9y^2+4x-6y\)
\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)
\(=\left(2x+1\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
\(d,x^2+4x-12\)
\(=x^2-2x+6x-12\)
\(=x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(x+6\right)\)
d,
\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-c\right)\)
Vậy..
e
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-2y-2\right)\left(x+2y\right)\)
a) x2-y2-2x+2y
=(x+y)(x-y)-2(x-y)
=(x-y)(x+y-2)
b) 2x + 2y - x2 -xy
=2(x+y) - x(x+y)
=(x+y)(2-x)
c) 3a2 - 6ab + 3b2 - 12c2
= 3(a2+b2) -3(2ab+4c2)
= 3(a2+b2-2ab-4c2)
d) x2 - 25 + y2 + 2xy
= x2 + 2xy + y2 -25
= (x+y)2 - 52
= (x+y+5)(x+y-5)
e) x2y - x3 - 9y + 9x
= (9x - x3)+(x2y -9y)
= x(9-x2)+y(x2 - 9)
= x(9-x2)-y(9-x2)
= (9-x2)(x-y)
f) x2-2x-4y2-4y
= x2-4y2-2(x+2y)
=(x+2y)(x-4y)-2(x+2y)
=(x+2y)(x-4y-2)
câu g trùng với câu e
h) x2(x-1)+16(1-x)
= x2(x-1)-16(x-1)
= (x2-16)(x-1)
= (x+4)(x-4)(x-1)