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a) \(x^2-x-y^2+y\)
\(=x\left(x-1\right)-y\left(y-1\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
b) \(x^2-2xy-z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
c) \(5x-5y+ax-ay\)
\(=5\left(x-y\right)+a\left(x-y\right)\)
\(=\left(x-y\right)\left(5+a\right)\)
x2-2xy+y2-z2+2zt-t2
=(x2-2xy+y2)-(z2-2zt+t2)
=(x-y)2-(z-t)2
=(x-y+z-t)(x-y-z+t)
x2 – 2xy + y2 – z2 + 2zt – t2 = (x2 – 2xy + y2) – (z2 – 2zt + t2)
= (x – y)2 – (z – t)2
= [(x – y) – (z – t)] . [(x – y) + (z – t)]
= (x – y – z + t)(x – y + z – t)
a) a3 - a2x - ay + xy
= (a3 - a2x) - (ay - xy)
= a2(a - x) - y(a - x)
= (a - x)(a2 - y)
b,c tương tự mà hình như b đề sai
1) \(25-x^2-y^2+2xy=5^2-\left(x^2-2xy+y^2\right)=5^2-\left(x-y\right)^2\)\(=\left(5-x+y\right)\left(5+x-y\right)\)
2) \(3x-3y-x^2+2xy-y^2\)\(=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)\(=3\left(x-y\right)-\left(x-y\right)^2\)\(=\left(x-y\right)\left(3-x+y\right)\)
1) \(25-x^2-y^2+2xy\)
\(=5^2-\left(x^2+y^2-2xy\right)\)
\(=5^2-\left(x-y\right)^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
2) \(3x-3y-x^2+2xy-y^2\)
\(=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)-\left(x-y\right)\left(x-y\right)\)
\(=\left(3-x+y\right)\left(x-y\right)\)
\(a,8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)
\(b,x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
\(c,4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)
1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
\(a,x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right).\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right).\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(b,9x^2+y^2+6xy=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
\(c,6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x^2-2.x.3+3^2\right)=-\left(x-3\right)^2\)
1) \(x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
2)\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-x\right)\left(x+y+z\right)\)
3)\(x^3+y^3-3x-3y=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-3\right)\)
\(1.x^3+y^3-x-y=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
2.\(3\left(x^2+6xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
3.\(\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-3\right)\)
cho mình nha
=\(x^2-2xy+y^2-\left(z^2-2zt+t^2\right)\))
=\(\left(x-y\right)^2\)- \(\left(z-t\right)^2\)