1) Phân tích đa thức thành nhân tử:
a) 6x2 – 9xy
b) x2 – 10x – 9y2 + 25
c) 3x2 – 3xy -2x + 2y
2) Chứng minh x2 – 6x + 10x > 0 với mọi số thực x.
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Bài 1:
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)
1, =3x (2x -3y)
c, = 3x(x-y) -2(x-y)
= (3x-2)(x-y)
2, Ta có: x2 -6x+10= (x-3)2 +11
Nhận xét: (x-3)2 >= 0 với mọi số thực x
=> (x-3)2 +1 >= 1 >0 (đpcm)
a: \(=\left(3-x\right)\left(x+1\right)\)
b: \(=3x\left(x-y\right)-5\left(x-y\right)\)
=(x-y)(3x-5)
c: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)
d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)
f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)
g) \(=y\left(y^2-2xy+x^2-y\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
a) 6x² + 7xy + 2y²
= 6x² + 4xy + 3xy + 2y²
= (6x² + 4xy) + (3xy + 2y²)
= 2x(3x + 2y) + y(3x + 2y)
= (3x + 2y)(2x + y)
b) x² - y² + 10x - 6y + 16
= x² + 10x + 25 - y² - 6y - 9
= (x² + 10x + 25) - (y² + 6y + 9)
= (x + 5)² - (y + 3)²
= (x + 5 - y - 3)(x + 5 + y + 3)
= (x - y + 2)(x + y + 8)
c) 4x⁴ + y⁴
= 4x⁴ + 4x²y² + y⁴ - 4x²y²
= (2x² + y²)² - (2xy)²
= (2x² + y² - 2xy)(2x² + y² + 2xy)
b) \(x^2y-x^3-10y+10x\)
\(=x^2\left(y-x\right)-10\left(y-x\right)\)
\(=\left(y-x\right)\left(x^2-10\right)\)
c) \(x^2\left(x-2\right)+49\left(2-x\right)\)
\(=\left(x-2\right)\left(x^2-49\right)\)
\(=\left(x-2\right)\left(x-7\right)\left(x+7\right)\)
a) \(3x^2-6xy=3x\left(x-2y\right)\)
b) \(x^3-6x^2+9x=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)
c) \(=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\)
d) \(=2x\left(3x-5\right)-3\left(3x-5\right)=\left(3x-5\right)\left(2x-3\right)\)
\(a,=3x\left(x-2y\right)\\ b,=x\left(x-3\right)^2\\ c,Sửa:x^2-2xy-3x+6y=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\\ d,=\left(3x-5\right)\left(2x-3\right)\)
a: =5x(x-y)-7(x-y)
=(x-y)(5x-7)
b: =x(x+2y)+(x+2y)
=(x+2y)(x+1)
c; =(x-3)^2-9y^2
=(x-3-3y)(x-3+3y)
a
\(5x^2-5xy+7y-7x\\ =5x\left(x-y\right)+7\left(y-x\right)\\ =5x\left(x-y\right)-7\left(x-y\right)\\ =\left(5x-7\right)\left(x-y\right)\)
b
\(x^2+2xy+x+2y\\ =x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+1\right)\left(x+2y\right)\)
c
\(x^2-6x-9y^2+9\\ =x^2-6x+9-\left(3y\right)^2\\ =x^2-2.x.3+3^2-\left(3y\right)^2\\ =\left(x-3\right)^2-\left(3y\right)^2\\ =\left(x-3-3y\right)\left(x-3+3y\right)\)
\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)
Xem lại đề ý b
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)