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x^2 - 7xy + 10y^2
= (x^2 - 2xy) - (5xy - 10y^2)
= x(x - 2y) - 5y( x - 2y)
= (x - 5y)(x - 2y)
\(x^5-3x^4-x^3-x^2+3x+1\)
\(=\left(x^5-x^2\right)-\left(3x^4-3x\right)-\left(x^3-1\right)\)
\(=x^2\left(x^3-1\right)-3x\left(x^3-1\right)-\left(x^3-1\right)\)
\(=\left(x^3-1\right)\left(x^2-3x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-\frac{3}{2}-\frac{\sqrt{13}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{13}}{2}\right)\)
\(x^5-3x^4-x^3-x^2+3x+1\)\(1\)\(=\left(x^5-x^4\right)-\left(2x^4-2x^3\right)-\left(3x^3-3x^2\right)-\left(4x^2-4x\right)-\left(x-1\right)\)
\(=x^4\left(x-1\right)-2x^3\left(x-1\right)-3x^2\left(x-1\right)-4x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4-2x^3-3x^2-4x-1\right)\)
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x^2+2\right)^2-y^4\)
\(=\left(x^2+y^2+2\right)\left(x^2-y^2+2\right)\)
\(\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
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hk tôt
a) x3 - 4x2 - 12x + 27
= \(\left(x^3+3x^2\right)-\left(7x^2+21x\right)+\left(9x+27\right)\)
= \(\left(x+3\right)\left(x^2-7x+9\right)\)
b) 9x2 + 6x - 8
=\(9x^2-6x+12x-8=3x\left(3x-2\right)+4\left(3x-2\right)\)
=\(\left(3x-2\right)\left(3x+4\right)\)
c) x2 - 7xy + 10y2
=\(x^2-5xy-2xy+10y^2=x\left(x-5y\right)-2y\left(x-5y\right)\)
=\(\left(x-5y\right)\left(x-2y\right)\)
a) x3 - 4x2 - 12x + 27
=x3 + 3x2 - 7x2 - 21x + 9x + 27
= x2(x+3) - 7x(x+3) + 9(x+3)
= (x2 - 7x + 9)(x + 3)
b) 9x2 + 6x - 8
= 9x2 - 6x + 12x - 8
= 3x(3x - 2) + 4(3x - 2)
= (3x + 4)(3x - 2)
c) x2 - 7xy + 10y2
= x2 - 5xy - 2xy + 10y2
= x(x - 5y) - 2y(x - 5y)
= (x - 2y)(x - 5y)
d) x8 + x7 + 1
Ta thêm vào các số hạng x6, x5, x4, x3, x2, x và cùng bớt đi các số hạng ấy ta có:
= x8 - x6 + x5 - x3 + x2 + x7 - x5 + x4 -x2 +x + x6 - x4 + x3 - x + 1
= x2(x6 - x4 + x3 - x + 1) + x(x6 - x4 + x3 - x + 1) + x6 - x4 + x3 - x + 1
= (x2 + x + 1)(x6 - x4 + x3 - x + 1)
i don't now
mong thông cảm !
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a/ \(=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x-7\right)\left(x+2\right)\)
b/ \(=15-3x=3\left(5-x\right)\)
c/ \(=x^2-3xy-4xy+12y^2=x\left(x-3y\right)-4y\left(x-3y\right)=\left(x-4y\right)\left(x-3y\right)\)
d/ \(=x^3-2x^2+x+2x^2-4x+2=x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=\left(x+2\right)\left(x-1\right)^2\)
a)x2-5x-14=(x2-7x)+(2x-14)=x(x-7)+2(x-7)=(x-7)(x+2)
b)4x2-3x-1=(4x2-4x)+(x-1)=4x(x-1)+(x-1)=(x-1)(4x+1)
c)x2-7xy+12y2=x2-7xy+12,25y2-0,25y2=(x-3,5y)2-0,25y2=(x-3,5y-0,5y)(x-3,5y+0,5y)=(x-4y)(x-2y)
d)x3-3x+2=(x3-x)-(2x-2)=(x-1)(x2+x)-2(x-1)=(x-1)(x2+x-2)=(x-1)(x2-2x+x-2)=(x-1)(x+1)(x-2)
1) \(3x^3-12x\)
\(=3x\left(x^2-4\right)\)
\(=3x\left(x+2\right)\left(x-2\right)\)
2)\(x^3+x^2-x-1\)
\(=\left(x^3+x^2\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-1\right)\)
\(=\left(x+1\right)^2\left(x-1\right)\)
3) \(x^2-7xy+10y^2\)
\(=\left(x^2-2xy\right)-\left(5xy-10y^2\right)\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-5y\right)\left(x-2y\right)\)
1) \(3x^3-12x\)
\(=3x\left(x^2-4\right)\)
\(=3x\left(x-2\right)\left(x+2\right)\)
2) \(x^3+x^2-x-1\)
\(=x^2\left(x+1\right)-\left(x+1\right)\)
\(=\left(x^2-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+1\right)\)