Phân tích đa thức thành nhân tử
\(2x^3-7x^2+x+10\)
phân tích đa thức thành nhân tử:
-2x^5-6x^4-8x^3-x^2+7x+10
phân tích đa thức sau thành nhân tử
-2x^4-7x^3-x^2+7x+3
\(-2x^4-7x^3-x^2+7x+3\)
\(=-2x^3\left(x+1\right)-5x^2\left(x+1\right)+4x\left(x+1\right)+3\left(x+1\right)\)
\(=-\left(x+1\right)\left(2x^3+5x^2-4x-3\right)\)
\(=-\left(x+1\right)\left[2x^2\left(x-1\right)+7x\left(x-1\right)+3\left(x-1\right)\right]\)
\(=-\left(x+1\right)\left(x-1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x+1\right)\left(x-1\right)\left(x+3\right)\left(2x+1\right)\)
phân tích đa thức thành nhân tử
\(a) x^4-7x^2+6\)
\(b) x^4+2x^2-3\)
\(c) x^3-2x^2+5x-4\)
a) \(=\left(x^2-6\right)\left(x^2-1\right)=\left(x^2-6\right)\left(x-1\right)\left(x+1\right)\)
b) \(=\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
c) \(=x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-x+4\right)\)
Phân tích đa thức thành nhân tử:
\(x^2+7x+10\)
\(=\left(x+2\right)\left(x+5\right)\)
\(x^2+7x+10=\left(x^2+5x\right)+\left(2x+10\right)=x\left(x+5\right)+2\left(x+5\right)=\left(x+2\right)\left(x+5\right)\)
1 phân tích đa thức thành nhân tử a. 7x^2-5x-2
b. x^3-7x^2-4x+10
2 tìm x biết 5.(2x-1)^2-3.(2x-1)=0
3 chứng minh x^2-4x+7>0
1. a) 7x2 - 5x - 2 = 7x2 - 7x + 2x - 2 = 7x(x - 1) + 2(x - 1) = (x - 1).(7x + 2)
2. 5(2x - 1)2 - 3(2x - 1) = 0
<=> (2x - 1).[5(2x - 1) - 3] = 0
<=> (2x - 1).(10x - 8) = 0
<=> (2x - 1) = 0 hoặc (10x - 8) = 0
<=> x = 1/2 hoặc x = 4/5
3. x2 - 4x + 7 = (x2 - 4x + 4) + 3 = (x - 2)2 + 3
Do: (x - 2)2 > hoặc = 0 (với mọi x)
Nên (x - 2)2 + 3 > hoặc = 3 (với mọi x)
Hay (x - 2)2 + 3 > 0 (với mọi x) => đpcm
\(7x^2-5x-2\)
\(=7x^2-7x+2x-2\)
\(=7x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(7x+2\right)\)
Phân tích đa thức thành nhân tử:
a) x^3+4x^2-7x-10
b) 8x^3(y+z)-y^3(z+2x)-z^3(2x-y)
Phân tích đa thức thành nhân tử
a/ \(5x^2-2x-3\)
b/ \(2x^2-3x-5\)
c/ \(x^2+2x-15\)
d/ \(7x^2-6x-1\)
\(a,=5x^2-5x+3x-3=\left(x-1\right)\left(5x+3\right)\\ b,=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\\ c,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ d,=7x^2-7x+x-1=\left(x-1\right)\left(7x+1\right)\)
c: =(x+5)(x-3)
d: =(x-1)(7x+1)
\(a,5x^2-2x-3=\left(5x^2-5x\right)+\left(3x-3\right)=5x\left(x-1\right)+3\left(x-1\right)=\left(x-1\right)\left(5x+3\right)\\ b,2x^2-3x-5=\left(2x^2+2x\right)-\left(5x+5\right)=2x\left(x+1\right)-5\left(x+1\right)=\left(x+1\right)\left(2x-5\right)\\ c,x^2+2x-15=\left(x^2-3x\right)+\left(5x-15\right)=x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(x+5\right)\\ d,7x^2-6x-1=\left(7x^2-7x\right)+\left(x-1\right)=7x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(7x+1\right)\)
phân tích đa thức thành nhân tử: 2x^4+7x^3-2x^2-13+6
\(2x^4+7x^3-2x^2-13x+6\)
\(=2x^4+6x^3+x^3+3x^2-5x^2-15x+2x+6\)
\(=2x^3\left(x+3\right)+x^2\left(x+3\right)-5x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(2x^3+x^2-5x+2\right)\left(x+3\right)\)
\(=\left(2x^3+4x^2-3x^2-6x+x+2\right)\left(x+3\right)\)
\(=\left[2x^2\left(x+2\right)-3x\left(x+2\right)+\left(x+2\right)\right]\left(x+3\right)\)
\(=\left(2x^2-3x+1\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(2x^2-2x-x+1\right)\left(x+2\right)\left(x+3\right)\)
\(=\left[2x\left(x-1\right)-\left(x-1\right)\right]\left(x+2\right)\left(x+3\right)\)
\(=\left(2x-1\right)\left(x-1\right)\left(x+2\right)\left(x+3\right)\)
Phân tích đa thức thành nhân tử dạng đoán nghiệm
a,-3x^4+20x^3-35x^2-10x+48
b,-2x^4-7x^3-x^2+7x+3
x^5-5x^4-2x^3+17x^2-13x+2
a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)
\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)
\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)
\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)
\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)
b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)
\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)
\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)
\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)