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KN
13 tháng 10 2019
a) \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4\right)^2-\left(7x\right)^2\)
\(=\left(12x-4\right)\left(-2x-4\right)\)
\(=-6\left(3x-1\right)\left(x+2\right)\)
c) \(x^2-y^2-x+y\)
\(=\left(x+y\right)\left(x-y\right)-\left(x-y\right)\)
\(=\left(x+y-1\right)\left(x-y\right)\)
d)\(4x^2-9y^2+4x-6y\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2y-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
KN
13 tháng 10 2019
e) \(-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
f) \(y^2\left(x^2+y\right)-zx^2-zy\)
\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
\(=\left(y^2-z\right)\left(x^2+y\right)\)
HV
1
TG
23 tháng 7 2020
a) \(y^2\left(x+y\right)-zx-zy\)
\(=y^2\left(x+y\right)-z\left(x+y\right)\)
\(=\left(x+y\right)\left(y^2-z\right)\)
b) \(x^2y+xy^2-x-y\)
\(=\left(x^2y-x\right)+\left(xy^2-y\right)\)
\(=x\left(xy-1\right)+y\left(xy-1\right)\)
\(=\left(xy-1\right)\left(x+y\right)\)
c) \(x^2+x-y^2+y\)
\(=\left(x^2-y^2\right)+\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+1\right)\)
d) \(x^3+x^2+x+1\)
\(=x^2\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+1\right)\)
18 tháng 3 2020
Ta có: a+b+c=1 <=>(a+b+c)2 = 1 <=> ab+bc+ca=0 (1)
Theo dãy tỉ số bằng nhau ta có:
xa=yb=zc=x+y+za+b+c=x+y+z1=x+y+zxa=yb=zc=x+y+za+b+c=x+y+z1=x+y+z
<=> x = a(x+y+z) ; y = b(x+y+z) ; z = c(x+y+z)
=> xy+yz+zx= ab(x+y+z)2+bc(x+y+z)2+ca(x + y + z)2
<=> xy+yz+zx =(ab+bc+ca)(x+y+z)2 (2)
từ (1) và (2) => xy + yz + zx = 0
25 tháng 5 2022
a: Sửa đề: \(y^2\left(x^2-y^2\right)-xz-yz\)
\(=y^2\left(x-y\right)\left(x+y\right)-z\left(x+y\right)\)
\(=\left(x+y\right)\left(xy^2-y^3-z\right)\)
c: \(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left(5x-5y\right)^2-\left(4x+4y\right)^2\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\cdot\left(9x-y\right)\)
TM
25 tháng 9 2017
a)\(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
b)\(x^2+11x+24=x^2+3x+8x+24=x\left(x+3\right)+8\left(x+3\right)=\left(x+3\right)\left(x+8\right)\)
c)\(xy\left(x-y\right)+y^2\left(y-z\right)+zx\left(z-x\right)=x\left[y\left(x-y\right)+z\left(z-x\right)\right]+y^2\left(y-z\right)\)
\(=x\left(xy-y^2+z^2-xz\right)+y^2\left(y-z\right)\)\(=x\left[\left(xy-xz\right)-\left(y^2-z^2\right)\right]+y^2\left(y-z\right)\)
\(=x\left[x\left(y-z\right)-\left(y-z\right)\left(y+z\right)\right]+y^2\left(y-z\right)\)\(=x\left(y-z\right)\left(x-y-z\right)+y^2\left(y-z\right)\)
\(=\left(y-z\right)\left(x^2-xy-xz\right)+y^2\left(y-z\right)=\left(y-z\right)\left(x^2-xy-xz+y^2\right)\)
22 tháng 9 2020
a) Ta có: \(\left(a-b\right)\left(a^2-c^2\right)-\left(a-c\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(a+c\right)-\left(a-c\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-c\right)\left(a-b\right)\left(a+c-a-b\right)\)
\(=\left(a-c\right)\left(a-b\right)\left(c-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
22 tháng 9 2020
b) Ta có: \(x^2-y^2+10x+8y+9\)
\(=\left(x^2+10x+25\right)-\left(y^2-8y+16\right)\)
\(=\left(x+5\right)^2-\left(y-4\right)^2\)
\(=\left(x+5-y+4\right)\left(x+5+y-4\right)\)
\(=\left(x-y+9\right)\left(x+y+1\right)\)