Mn nguoi giúp vs ak.
6x^3-7x^2+5x-2
4x^3+5x^2+10x-12
4x^3-7x^2-x+3
4x^3-5x^2+6x+9
x^3-12x^2+14x-4
3x^3-5x^2+5x-2 mn phân tích bằng phường pháp hệ so bất dinh nha
Bài 1
a) 2x^3 + 5x^2 + 5x + 3
b) 4x^3 + x^2 + x - 3
c) 5x^3 - 12x^2 + 14x - 4
d) 6x^3 - 7x^2 + 5x - 2
e) 3x^3 + 19x^2 + 4x - 12
\(a,2x^3+5x^2+5x+3\)
\(=2x^3+3x^2+2x^2+3x+2x+3\)
\(=x^2\left(2x+3\right)+x\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+x+1\right)\)
b) = 4x^3 - 3x^2 + 4x^2 - 3x + 4x - 3
= x^2(4x-3) + x(4x - 3) + 4x - 3
= (4x - 3)(x^2 + x + 1)
c) = 5x^3 - 2x^2 - 10x^2 + 4x + 10x - 4
= x^2(5x - 2) - 2x(5x - 2) + 2(5x - 2)
= (5x - 2)(x^2 - 2x + 2)
d)= 6x^3 - 4x^2 - 3x^2 + 2x + 3x - 2
= 2x^2(3x - 2) - x(3x - 2) + (3x - 2)
= (3x-2)(2x^2-x+1)
e) = 3x^3 - 2x^2 + 21x^2 - 14x + 18x - 12
= x^2( 3x - 2) + 7x(3x - 2) + 6(3x - 2)
= (3x - 2)(x^2 + 7x + 6)
= (3x - 2)(x+1)(x+6)
x^2-5x+6
x^2-7x+12
x^2+x-12
x^2-9x+20
2x^2-3x-2
4x^2-7x-2
4x^2+15x+9
\(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
\(x^2-7x+12=\left(x-2\right)\left(x-5\right)\)
\(x^2+x-12=\left(x-5\right)\left(x+6\right)\)
\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)
\(2x^2-3x+2=2\left(x+\dfrac{1}{2}\right)\left(x-2\right)\)
\(4x^2-7x-2=4\left(x-2\right)\left(x+\dfrac{1}{4}\right)\)
\(4x^2+15x+9=4\left(x+\dfrac{3}{4}\right)\left(x+3\right)\)
Phân tích thành nhân tử ( tách các hạng tử)
1,x^2-6x+3
2, 2m^2+10m+8
3, 9x^2+6x-8
4, x^3-5x^2-14x
5, a^4+4a^2-5
6, x^3-7x-6
7, 3x^2-22xy+4x+8y+7y^2+1
8, 12x^2+5x-12y^2+12y-10xy-3
1) \(x^2-6x+3\)
\(=x^2-6x+9-6\)
\(=\left(x-3\right)^2-6\)
\(=\left(x-3+\sqrt{6}\right)\left(x-3-\sqrt{6}\right)\)
2) \(2m^2+10m+8\)
\(=2m^2+2m+8m+8\)
\(=2m\left(m+1\right)+8\left(m+1\right)\)
\(=\left(2m+8\right)\left(m+1\right)\)
\(=2\left(m+4\right)\left(m+1\right)\)
3) \(9x^2+6x-8\)
\(=\left(9x^2+6x+1\right)-9\)
\(=\left(3x+1\right)^2-9\)
\(=\left(3x+4\right)\left(3x-2\right)\)
4) \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-2x+7x-14\right)\)
\(=x\left[x\left(x-2\right)+7\left(x-2\right)\right]\)
\(=x\left(x+7\right)\left(x-2\right)\)
1) x^2-6x+3
= (x^2-6x+9)-6
=(x-3)^2-6
=(x-3-căn 6)(x-3+căn 6)
2) =2(m^2+5m+4)
=2(m+1)(m+4)
3) =9x^2+6x+1-9
=(3x+1)^-9
=(3x-2)(3x+4)
4, x^3-5x^2-14x
=x(x-7)(x+2)
5, a^4+4a^2-5
=a^4+4a^2+4-9
=(a^2+2)^-9
=(a^2-1)(a^2+5)
6, x^3-7x-6
=(x-3)(x+1)(x+2).
5) \(a^4+4a^2-5\)
\(=a^4+5a^2-a^2-5\)
\(=a^2\left(a^2+5\right)-\left(a^2+5\right)\)
\(=\left(a^2-1\right)\left(a^2+5\right)\)
\(=\left(a+1\right)\left(a-1\right)\left(a^2+5\right)\)
6) \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x^2+2x-3x-6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\)
Bài 1
a) 2x^3 + 5x^2 + 5x + 3
b) 4x^3 + x^2 + x - 3
c) 5x^3 - 12x^2 + 14x - 4
d) 6x^3 - 7x^2 + 5x - 2
e) 3x^3 + 19x^2 + 4x - 12
phân tích đa thức thành nhân tử
3x^2+10x+3
5x^2+14x-3
6x^2+7x-3
3x2 + 10x +3
=> 3x2 + 9x +x +3
=> (3x2 +x )+ ( 9x +3)
=> x(3x +1) +3( 3x+1)
=>(3x+1) (x+3)
Tự làm tiếp nhé
a, ta có: 3x^2 +10x+3= 3x^2 + 9x +x +3 = 3x(x+3)+(x+3)=(x+3)+(3x+1)
2 câu còn lại đang trong qtrinhf suy nghĩ
a) \(3x^2+10x+3=3x^2+9x+x+3\)
\(=3x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(3x+1\right)\)
b)\(5x^2+14x-3=5x^2+15x-x-3\)
\(=5x\left(x+3\right)-\left(x+3\right)\)
\(=\left(5x-1\right)\left(x+3\right)\)
c)\(6x^2+7x-3=6x^2+9x-2x-3\)
\(=3x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(3x-1\right)\left(2x+3\right)\)
phân tích đa thức thành nhân tử
a, 2x^4-5x^3-27x^2 + 25x + 50
b, 3x^4 + 6x^3- 33x^2-24x+48
c, x^4+ 7x^3+14x^2+14x+4
b) 3x4-3x3+9x3-9x2-24x2+24x-48x+48
=3x3(x-1)+9x2(x-1)-24x(x-1)-48(x-1)
=(x-1)(3x3+9x2-24x-48)
=3(x-1)(x3+3x2-8x-16)
d) (5x+3) ( 4x-1) +(10x-7) (-2x+3) =27
e)(8x-5) (3x+2) -(12x+7) (2x-1)=17
f) (5x+9) (6x-1) -(2x-3)( 15z+1) = -190
g) 6x(5x+3) + 3x(1-10x) =7
h) (3x-3) (5 -21x) +(7x+4)(9x-5) =44\
i) (x+1)(x+2)(x-5)-x2 (x+8)=27
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
phân tích thành nhân tử :
1, \(4x^3+5x^2+10x+12\)
2, \(6x^3-7x^2+5x-2\)
3, \(4x^3-x^2+x+3\)
4, \(3x^3-5x^2+5x-2\)
5, \(5x^3-12x^2+14x-4\)
4: \(3x^3-5x^2+5x-2\)
\(=3x^3-2x^2-3x^2+2x+3x-2\)
\(=x^2\left(3x-2\right)-x\left(3x-2\right)+\left(3x-2\right)\)
\(=\left(3x-2\right)\left(x^2-x+1\right)\)
5: \(5x^3-12x^2+14x-4\)
\(=5x^3-2x^2-10x^2+4x+10x-4\)
\(=\left(5x-2\right)\left(x^2-2x+2\right)\)
Phân tích đa thức thành nhân tử bằng phương pháp hệ số bất định:
a. 3x^4+11x^3-7x^2-2x+
b. x^4-6x^3+11x^2-6x+1
c. x^4-x^3+2x^2-11x-5
e. x^4++6x^3+11x^2+6x+1
f. 4x^4+4x^3+5x^2+5x+1
g. x^4-7x^3+14x^2-7x+1
Giúp mk vs. Ý nào cx đc
câu c:x^4-2x^3-x^2+x^3-2x^2-x+5x^2-10x-5=x^2(x^2-2x-1)+x(x^2-2x-1)+5(x^2-2x-1)=(x^2-2x-1)(x^2+x+5)