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a) 3x2 - 11x + 6
= 3x2 - 9x - 2x + 6
= 3x(x - 3) - 2(x - 3)
= (x - 3) (3x - 2)
b) 8x2 + 10x - 3
= 8x2 - 2x + 12x - 3
= 2x(4x - 1) - 3(4x - 1)
= (4x - 1) (2x - 3)
c) 8x2 - 2x - 1
= bó tay!!!
ok mk nha!!! 565464565657565878797624534534344565464513535465656756
\(a,3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
\(b,8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
\(c,8x^2-2x-1=9x^2-x^2-2x-1=9x^2-\left(x+1\right)^2=\left(3x-x-1\right)\left(3x+x+1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
a.\(3x^2-11x+6\)
= \(3x^2-9x-2x+6\)
=\(3x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-3\right)\left(3x-2\right)\)
b\(8x^2+10x-3\)
=.\(8x^2-2x+12x-3\)
=\(2x\left(4x-1\right)+3\left(4x-1\right)\)
=\(\left(4x-1\right)\left(2x+3\right)\)
d.\(x^2-y^2+10x-6y+16\)
=\(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)
=\(\left(x+5\right)^2-\left(y+3\right)^2\)
=\(\left(x+5-y-3\right)\left(x+5+y+3\right)\)
=\(\left(x-y+2\right)\left(x+y+8\right)\)
e.\(x^4+x^2y^2+y^4\)
=\(x^4+2x^2y^2+y^4-x^2+y^2\)
=\(\left(x^2+y^2\right)^2-x^2y^2\)
=\(\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
a)
\(=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
a) ta có 3x^2 -11x+6= 3x^2 - 9x -2x + 6 = 3x (x-3) - 2(x-3) =(x-3) (3x-2)
b) ta có 8x^2 + 10x - 3 = 2x (4x-1) - 3(4x- 1) = (2x-3)(4x-1)
c) ta có 8x^2 -2x - 1 = 8x^2 - 4x +2x-1 = 4x(2x-1) + 2x- 1 = (4x+1)(2x-1) k nha bạn
a, 3x2 - 11x+6 = 3x2 - 9x - 2 x + 6 = 3x(x-3) - 2(x-3) = (3x-2)(x-3)
3x-2 = 0 và x - 3 =0
x = 2/3 x = 3
a, \(3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
b. \(8x^2-2x+12x-3=2x\left(4x-1\right)+3\left(4x-1\right)=\left(4x-1\right)\left(2x+3\right)\)
c. đề kiểu gì vậy? -2x-x để thành -3x à? xem lại đi nha
d. \(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+5-y-3\right)\left(x+5+y+3\right)=\left(x-y+2\right)\left(x+y+8\right)\)
e. \(=x^4+2x^2y^2+y^4-x^2y^2=\left(x^2+y^2\right)^2-x^2y^2=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
nhớ L I K E
1 (x-3)(3x-2)
2 (4x-1)(2x+3)
câu 3 bị sai thì phải bạn cứ nhân ra thì biết cách phân tích thôi mà
phân tích đa thức ->nhân tử:
a)2x2+4x-70
b)x3-5x2+8x-4
c)x2-10+16
rút gọn:
(8x-8x3-10x2+3x4-5):(3x2-2x+1)
Bài 1:
a)2x2+4x-70
=2(x2+2x-35)
=2(x2+7x-5x-35)
=2[x(x+7)-5(x+7)]
=2(x-5)(x+7)
b)x3-5x2+8x-4
=x3-4x2+4x-x2+4x-4
=x(x2-4x+4)-(x2-4x+4)
=(x2-4x+4)(x-1)
=(x-2)2(x-1)
c)x2-10x+16
=x2-2x-8x+16
=x(x-2)-8(x-2)
=(x-8)(x-2)
Bài 2:
\(\frac{8x-8x^3-10x^2+3x^4-5}{3x^2-2x+1}=\frac{\left(x^2-2x-5\right)\left(3x^2-2x+1\right)}{3x^2-2x+1}=x^2-2x-5\)
a, \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\text{[}x\left(x+1\right)+2\left(x+1\right)\text{]}\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
b, \(2x^3+3x^2+3x+2\)
\(=2x^3+2x^2+x^2+x+2x+2\)
\(=2x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2+x+2\right)\)
c, \(x^3-4x^2-8x+8\)
\(=x^3+2x^2-6x^2-12x+4x+8\)
\(=x^2\left(x+2\right)-6x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-6x+4\right)\)
a)
\(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
b)
\(8x^2+10x-3\)
\(=8x^2-2x+12x-3\)
\(=8x\left(x-\frac{1}{4}\right)+12\left(x-\frac{1}{4}\right)\)
\(=\left(8x+12\right)\left(x-\frac{1}{4}\right)\)
\(=\left(2x+2\right)\left(4x-1\right)\)
c)
\(8x^2-2x-1\)
\(=8x^2-4x+2x-1\)
\(=8x\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{2}\right)\)
\(=\left(x-\frac{1}{2}\right)\left(8x+2\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
a. 3x^2-11x + 6
= 3x^2 - 9x - 2x +6
= 3x(x-3) - 2(x-3)
= (x-3)(3x-2)