Phân tích đa thức thành nhân tử bằng cách tách một số hạng tử thành nhiều số hạng khác
a) x2 - 13 + 36
b) x2 + 3x - 18
c) x2 - 5x - 24
d) 3x2 - 16x + 5
e) 8x2 + 30 + 7
g) 2x2 - 5x - 12
h) 6x2 - 7x - 20
i) x4 + 4
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\(1,2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+\left(x-2\right)\)
\(=\left(2x+1\right)\left(x-2\right)\)
\(2,4x^2-7x-2\)
\(=4x^2-8x+x-2\)
\(=4x\left(x-2\right)+x-2\)
\(\left(4x+1\right)\left(x-2\right)\)
a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x-6y-1\right)\)
b) \(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c) \(=2\left(x-y\right)^2-18\)
\(=2\left[\left(x-y\right)^2-3^2\right]\)
\(=2\left(x-y+3\right)\left(x-y-3\right)\)
a: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: \(x^3-8x^2+16x\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
a: Ta có: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: Ta có: \(16x-8x^2+x^3\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: Ta có: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\cdot\left[\left(x-y\right)^2-9\right]\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: Ta có: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
e: Ta có: \(x^4-x^2-30\)
\(=x^4-6x^2+5x^2-30\)
\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)
\(=\left(x^2-6\right)\left(x^2+5\right)\)
f: Ta có: \(x^2-xy-2y^2\)
\(=x^2-2xy+xy-2y^2\)
\(=x\left(x-2y\right)+y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+y\right)\)
g: Ta có: \(x^4-13x^2y^2+4y^4\)
\(=x^4-4x^2y^2+4y^4-9x^2y^2\)
\(=\left(x^2-2y^2\right)^2-\left(3xy\right)^2\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-3xy+2y^2\right)\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-xy-2xy+2y^2\right)\)
\(=\left[x\left(x-y\right)-2y\left(x-y\right)\right]\left(x^2-3xy-2y^2\right)\)
\(=\left(x-y\right)\left(x-2y\right)\left(x^2-3xy-2y^2\right)\)
h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)^2-3\left(x^2-2x\right)+\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x^2-2x+1\right)\)
\(=\left(x-3\right)\left(x+1\right)\cdot\left(x-1\right)^2\)
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
a: =(x+6)(x-1)
n: \(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)
f)\(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i)\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-5\right)\left(x-2\right)\)
h)\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)
g)\(x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
f)\(x^2-5x-14=x^2-7x+2x-14\)
\(=\left(x+2\right)\left(x-7\right)\)
i)\(x^2-7x+10=x^2-5x-2x+10\)
\(=\left(x-2\right)\left(x-5\right)\)
h)\(x^2-7x+12=x^2-4x-3x+12\)
\(=\left(x-3\right)\left(x-4\right)\)
g)\(x^2+6x+5=x^2+x+5x+5\)
\(=\left(x+5\right)\left(x+1\right)\)
a,x2 - 13x + 36
= x2 - 4x - 9x + 36
= x(x - 4) - 9(x - 4)
= (x - 4)(x - 9)
b,x2 + 3x - 18
= x2 - 9 + 3x - 9
= (x - 3)(x + 3) + 3(x - 3)
= (x - 3)(x + 3 + 3)
= (x - 3)(x + 6)
c,x2 - 5x - 24
= x2 + 3x - 8x - 24
= x(x + 3) - 8(x + 3)
= (x + 3)(x - 8)
d,3x2 - 16x + 5
= 3x2 - 15x - x + 5
= 3x(x - 5) - (x - 5)
= (x - 5)(3x - 1)
e, 8x2 + 30x + 7
= 8x2 + 2x + 28x + 7
= 2x(4x + 1) + 7(4x + 1)
= (4x + 1)(2x + 7)
g,2x2 - 5x - 12
= 2x2 - 8x + 3x - 12
= 2x(x - 4) + 3(x - 4)
= (x - 4)(2x + 3)
i,x4 + 4
= x4 + 4x2 + 4 - 4x2
= (x2 + 2)2 - (2x)2
= (x2 + 2 - 2x)(x2 + 2 + 2x)
Phân tích đa thức thành nhân tử bằng cách tách một số hạng tử thành nhiều số hạng khác
a) x^2+4x+3
b) 4x^2-4x-3
c)x^2-x-12
d) 4x^4-4x^2-8y^4