Phân tích đa thức thành nhân tử ( phương pháp tách hạng tử)
1) 2x2 + 3x - 5
2) 16x - 5x2 - 3
3) 7x - 6x2 - 2
4) x2 + 5x - 6
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1/ \(x^2+x-90=\left(x^2-10x\right)+\left(9x-90\right)=x\left(x-10\right)+9\left(x-10\right)=\left(x-10\right)\left(x+9\right)\)
2/ \(2x^2+4xy+2y^2=\left(2x^2+2xy\right)+\left(2xy+2y^2\right)=2x\left(x+y\right)+2y\left(x+y\right)=\left(x+y\right)\left(2x+2y\right)\)
3/ \(2y^2-14y+24=2\left(y^2-7y+12\right)=2\left[\left(y^2-4y\right)+\left(12-3y\right)\right]=2\left[y\left(y-4\right)-3\left(y-4\right)\right]\)
\(=2\left(y-4\right)\left(y-3\right)\)
4/ \(x^8+x^4+1=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x^6-x^5+x^4\right)-\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^4\left(x^2-x+1\right)\right]-x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\)
1) x2 - 4x + 3
= x2 - x - 3x + 3
= (x2 - x) - (3x - 3)
= x.(x - 1) - 3.(x - 1)
= (x - 1).(x - 3)
2) x2 - x - 6
= x2 + 2x - 3x - 6
= (x2 + 2x) - (3x + 6)
= x.(x + 2) - 3.(x + 2)
= (x + 2).(x - 3)
3) x2 + 5x + 4
= x2 + x + 4x + x
= (x2 + x) + (4x + x)
= x.(x + 1) + 4.(x + 1)
= (x + 1).(x + 4)
4) x2 + 5x + 6
= x2 + 2x + 3x + 6
= (x2 + 2x) + (3x + 6)
= x.(x + 2) + 3.(x + 2)
= (x + 2).(x + 3)
a,=x^2+x+3x+3
=x(x+1)+3(x+1)
=(x+3)(x+1)
b,x^2-3x+2x-6
=x(x-3)+2(x-3)
=(x+2)(x-3)
2 câu còn lại từ lm nha.........
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) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
\(1,2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+\left(x-2\right)\)
\(=\left(2x+1\right)\left(x-2\right)\)
\(2,4x^2-7x-2\)
\(=4x^2-8x+x-2\)
\(=4x\left(x-2\right)+x-2\)
\(\left(4x+1\right)\left(x-2\right)\)
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
a) \(x^3+4x^2-21x\)
\(=x\left(x^2+4x-21\right)\)
\(=x\left(x^2-3x+7x-21\right)\)
\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)
\(=x\left(x-3\right)\left(x+7\right)\)
b) \(5x^3+6x^2+x\)
\(=x\left(5x^2+6x+1\right)\)
\(=x\left(5x^2+5x+x+1\right)\)
\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(5x+1\right)\)
c) \(x^3-7x+6\)
\(=x^3+2x^2-3x-2x^2-4x+6\)
\(=x\left(x^2+2x-3\right)-2\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+3\right)\)
d) \(3x^3+2x-5\)
\(=3x^3+3x^2+5x-3x^2-3x-5\)
\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)
\(=\left(x-1\right)\left(3x^2+3x+5\right)\)
1) \(25-x^2-y^2+2xy=5^2-\left(x^2-2xy+y^2\right)=5^2-\left(x-y\right)^2\)\(=\left(5-x+y\right)\left(5+x-y\right)\)
2) \(3x-3y-x^2+2xy-y^2\)\(=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)\(=3\left(x-y\right)-\left(x-y\right)^2\)\(=\left(x-y\right)\left(3-x+y\right)\)
1) \(25-x^2-y^2+2xy\)
\(=5^2-\left(x^2+y^2-2xy\right)\)
\(=5^2-\left(x-y\right)^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
2) \(3x-3y-x^2+2xy-y^2\)
\(=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)-\left(x-y\right)\left(x-y\right)\)
\(=\left(3-x+y\right)\left(x-y\right)\)
e) \(8\left(x+3y\right)-16x\left(x+3y\right)=\left(x+3y\right)\left(8-16x\right)=8\left(x+3y\right)\left(1-2x\right)\)
f) \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=\left(x+1\right)\left(4x^2+2x^2\right)=6x^2\left(x+1\right)\)
g) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)
a)3x2+7x-6
=3x2-2x+9x-6
=x(3x-2)+3(3x-2)
=(x+3)(3x-2)
b)8x2-2x-3
=8x2-6x+4x-3
=2x(4x-3)+(4x-3)
=(2x+1)(4x-3)
c)6x2-15x+6
=3(2x2-5x+2)
=3(2x2-x-4x+2)
=3[x(2x-1)-2(2x-1)]
=3(x-2)(2x-1)
d)10x2+7x-6
=10x2+12x-5x-6
=2x(5x+6)-(5x+6)
=(2x-1)(5x+6)
1/ \(2x^2+3x-5=\left(2x^2+2x\right)-\left(5x+5\right)=2x\left(x+1\right)-5\left(x+1\right)=\left(x+1\right)\left(2x-5\right)\)
2/ \(16x-5x^2-3=\left(15x-5x^2\right)+\left(x-3\right)=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)
3/ \(7x-6x^2-2=\left(3x-6x^2\right)-\left(2-4x\right)=3x\left(1-2x\right)-2\left(1-2x\right)=\left(1-2x\right)\left(3x-2\right)\)
4/ \(x^2+5x-6=\left(x^2-x\right)+\left(6x-6\right)=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)