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\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(---\)
\(x^2-16y^2+4x+4\)
\(=\left(x^2+4x+4\right)-16y^2\)
\(=\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left(x+2-4y\right)\left(x+2+4y\right)\)
\(=\left(x-4y+2\right)\left(x+4y+2\right)\)
\(---\)
\(3x^2+6xy+3y^2-12\)
\(=3\left(x^2+2xy+y^2-4\right)\)
\(=3\left[\left(x+y\right)^2-2^2\right]\)
\(=3\left(x+y-2\right)\left(x+y+2\right)\)
\(---\)
\(4x^3+4x^2+x\)
\(=x\left(4x^2+4x+1\right)\)
\(=x\left(2x+1\right)^2\)
a) \(xy+y^2-x-y\)
\(=\left(xy+y^2\right)-\left(x+y\right)\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
a) xy +y2 - x-y
y(x+y) -(x+y)
(x+y)(y-1)
c) x2 - 4x +3
x2 -3x - x - 3
x(x-3) -(x-3)
(x-3)(x-1)
câu 2
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ĐỂ phép chia hết thì m+12 = 0 => m = -12
có thể đúng cũng có thể sai ,có j sai hoặc ko đúng ib mk nhé
\(x^2-y^2+3x-3y\)
\(=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3.\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
\(x^2-y^2+4x+4\)
\(=\left(x^2+2.2x+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
Tham khảo nhé~
\(x^3+4x^2+4x-16y^2\)
\(=\left(x^3+2x^2\right)+\left(2x^2+4x\right)-16y^2\)
\(=x^2.\left(x+2\right)+2x.\left(x+2\right)-16y^2\)
\(=\left(x+2\right).\left(x^2+2x\right)-16y^2\)
\(=x.\left(x+2\right).\left(x+2\right)-\left(4y\right)^2\)
\(=x.\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left[\sqrt{x}.\left(x+2\right)\right]^2-4y^2\)
\(=\left[\sqrt{x}.\left(x+2\right)-4y\right].\left[\sqrt{x}.\left(x+2\right)+4y\right]\)
Tham khảo nhé~
nếu đưa vô căn phải có điều kiện là x > 0
\(x^3+4x^2+4x-16y^2=x\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left(x\sqrt{x}+2\sqrt{x}\right)^2-\left(4y\right)^2=\left(x\sqrt{x}+2\sqrt{x}-4y\right)\left(x\sqrt{x}+2\sqrt{x}+4y\right)\)
\(1,\)
\(x^2-2x-4y^2-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\(2,\)
\(x^4+2x^3-4x-4\)
\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
\(3,\)
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)[3\left(x+y\right)-2\left(x-y\right)]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
\(4,\)
\(x^2-y^2-2x+2y\)
\(=x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(a,=x\left(y-3\right)+y\left(y-3\right)=\left(x+y\right)\left(y-3\right)\\ b,=\left(x+2\right)^2-16y^2=\left(x+4y+2\right)\left(x-4y+2\right)\)
\(a,xy-3x+y^2-3y=\left(xy-3x\right)+\left(y^2-3y\right)=x\left(y-3\right)+y\left(y-3\right)=\left(x+y\right)\left(y-3\right)\\ b,x^2-16y^2+4x+4=\left(x^2+4x+4\right)-16y^2=\left(x+2\right)^2-\left(4y\right)^2=\left(x-4y+2\right)\left(x+4y+2\right)\)