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Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
\(a^4+8a^3+14a^2-8a-15\)
\(=a^4+8a^3+15a^2-a^2-8a-15\)
\(=a^2\left(a^2+8a+15\right)-\left(a^2+8a+15\right)\)
\(=\left(a^2+8a+15\right)\left(a^2-1\right)\)
\(=\left(a+3\right)\left(a+5\right)\left(a-1\right)\left(a+1\right)\)
\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right).\left[x^2+4x-3x-12\right]\)
\(=\left(x-2\right).\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
\(x^4+x^3+2x-4\)
\(=x^4-x^3+2x^3-2x^2+2x^2-2x+4x-4\)
\(=x^3\left(x-1\right)+2x^2\left(x-1\right)+2x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+2x+4\right)\)
\(=\left(x-1\right).\left[x^2\left(x+2\right)+2\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+2\right)\)
\(8x^4-2x^3-3x^2-2x-1\)
\(=8x^4-8x^3+6x^3-6x^2+3x^2-3x+x-1\)
\(=8x^3\left(x-1\right)+6x^2\left(x-1\right)+3x\left(x-1\right)+x-1\)
\(=\left(x-1\right)\left(8x^3+6x^2+3x+1\right)\)
\(=\left(x-1\right)\left[\left(8x^3+1\right)+\left(6x^2+3x\right)\right]\)
\(=\left(x-1\right)\left[\left(2x+1\right)\left(4x^2-2x+1\right)+3x\left(2x+1\right)\right]\)
\(=\left(x-1\right)\left(2x+1\right)\left(4x^2+x+1\right)\)
\(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
Chúc bạn học tốt.
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
=x3(x+2)-13x2+12x-26x+24
=x3(x+2)-x(13x-12)-2(13x-12)
=x3(x+2)-(13x-12)(x+2)
=(x+2)(x3-x-12x+12)
(x+2)[(x2-1)-12(x-1)]
=(x+2)[x(x-1)(x+1)-12(x-1)]
=(x+2)(x-1)[x(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x(x-3)+4(x+3)]
=(x+2)(x-1)(x-3)(x+4)
trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!
2.,
A = \(3x^2+2x-1=3\left(x^2+\frac{2}{3}x-\frac{1}{3}\right)=3\left(x^2+\frac{2.x.1}{3}+\frac{1}{9}-\frac{1}{9}-\frac{1}{3}\right)\)
A = \(3\left[\left(x+\frac{1}{3}\right)^2-\frac{4}{9}\right]=3\left(x+\frac{1}{3}\right)^2-\frac{4}{3}\)
VẬy GTNN của A là -4/3 khi x = -1/3 ( GTNN không có GTLN đâu nha)
B = \(-9x^2+3x=-\left(9x^2-3x\right)=-\left(9x^2-2.3x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)\)
B = \(-\left(3x+\frac{1}{2}\right)^2+\frac{1}{4}\)
VẬy GTLN của B = 1/4 khi 3x + 1/2 = 0
2x3y - 2xy3 - 4xy2 - 2xy
= 2xy (x2 - y2 - 2y - 1)
= 2xy [x2 - (y2 + 2y + 1)]
= 2xy [x2 - (y + 1)2]
= 2xy (x - y - 1) (x + y + 1)
\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
A . 5(x-y)-y(x-y)
=(x6-y)(5-y)
B . x^2 - xy - 8x+8y
=(x^2-xy)-(8x-8y))
=x(x-y) - 8(x-y)
C. x^2-10x+25 - y^2
=(x^2 - 10x + 25 ) - y^2
=(x-5)^2 - y^2
=(x-5+y)(x-5-y)
D . x^3 - 3x^2-4x+12
=(x^3 - 3x^2 ) - (4x - 12)
=x^2 (x-3)-4(x-3)
=(x^2-4)(x-3)
=(x+2)(x-2)(x-3)
D . 2x^2-2y^2- 6x-6y
=(2^x - 2y^2) - (6x+ 6y)
=2(x^2 - y^2) - 6(x+y)
=2(x+y)(x-y) - 6(x+y)
=2(x+y)(x-y-3)
E . x^3 - 3x^2 + 3x - 1
=(x-1)^3
D.x^2+3x+2
=x^2+2x+x+2
=(x^2+2x)+(x+2)
=x(x+2)+(x+2)
=(x+2)(x+1)
`2x^3-3x^2-32x-15`
`=2x^3-10x^2+7x^2-35x+3x-15`
`=2x^2(x-5)+7x(x-5)+3(x-5)`
`=(x-5)(2x^2+7x+3)`
`=(x-5)(2x^2+x+6x+3)`
`=(x-5)[x(2x+1)+3(2x+1)]`
`=(x-5)(2x+1)(x+3)`