Bài 1:
a) \(ay-ax-2x+2y\)
\(=-a\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(-a-2\right)\)
b) \(5ax-7by-7ay+5bx\)
\(=5x\left(a+b\right)-7y\left(a+b\right)\)
\(=\left(a+b\right)\left(5x-7y\right)\)
c) \(4x^2-9x+5\)
\(=4x^2-4x-5x+5\)
\(=4x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(4x-5\right)\)
d) \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
Bài 2:
a) \(x^2+x+\frac{1}{2}\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{1}{4}>0\forall x\)
b) \(x^2+5x+7\)
\(=x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}+\frac{3}{4}\)
\(=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}>0\forall x\)
c) \(2x^2-3x+9\)
\(=2\left(x^2-\frac{3}{2}x+\frac{9}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}+\frac{63}{16}\right)\)
\(=2\left[\left(x-\frac{3}{4}\right)^2+\frac{63}{16}\right]\)
\(=2\left(x-\frac{3}{4}\right)^2+\frac{63}{8}>0\forall x\)
Bài 1: Phân tích đa thức thành nhân tử.
a, ay - ax - 2x + 2y
=a(y-x)+2(y-x)=(y-x)(a+2
b, 5ax - 7by - 7ay + 5bx
=5x(a+b)-7y(b+a)=(a+b)(5x-7y)
c, 4x^2 - 9x + 5
=4x2-4x-5x+5=4x(x-1)-5(x-1)=(x-1)(4x-5)
d, x^2 - 8x + 15
=x2-3x-5x+15=x(x-3)-5(x-3)=(x-3)(x-5)
Bài 2: Chứng minh rằng.
a, x^2 + x + 1/2 >0 với mọi x (mình nghĩ phải là >0 )
Ta có :
x2+x+1/2=x2+2x.1/2+1/4 +1/4=(x+1/2)2+1/4
vì (x+1/2)2 ≥0 ∀ x nên (x+1/2)2+1/4 ≥ 1/4>0 ∀ x
Vậy x^2 + x + 1/2 >0 với mọi x
