Đề thiếu. Bạn coi lại đề.

\(P=\dfrac{a^2}{ab+b^2}+\dfrac{b^2}{ab-a^2}-\dfrac{a^2+b^2}{ab}\) (\(a\ne b;a\ne0;a\ne-b;b\ne0\))

\(=\dfrac{a^2}{b\left(a+b\right)}+\dfrac{b^2}{a\left(b-a\right)}-\dfrac{a^2+b^2}{ab}\)

\(=\dfrac{a^3\left(a-b\right)-b^3\left(a+b\right)-\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)}{ab\left(a+b\right)\left(a-b\right)}\)

\(=\dfrac{a^4-a^3b-b^3a-b^4-\left(a^2+b^2\right)\left(a^2-b^2\right)}{ab\left(a+b\right)\left(a-b\right)}\)

\(=\dfrac{a^4-a^3b-b^3a-b^4-\left(a^4-b^4\right)}{ab\left(a+b\right)\left(a-b\right)}\)

\(=\dfrac{-a^3b-b^3a}{ab\left(a+b\right)\left(a-b\right)}\)

\(=\dfrac{-ab\left(a^2+b^2\right)}{ab\left(a+b\right)\left(a-b\right)}=-\dfrac{a^2+b^2}{a^2-b^2}\).

b) -Ta có: \(P=0\)

\(\Leftrightarrow-\dfrac{a^2+b^2}{a^2-b^2}=0\)

\(\Leftrightarrow a^2+b^2=0\)

-Vì \(a^2\ge0;b^2\ge0\)

\(\Rightarrow a=0;b=0\) (không thỏa mãn điều kiện).

-Vậy không có giá trị nào của a,b để \(P=0\).

c) 

a,x²-y²-2x+2y
= (x+y)(x-y) - 2(x-y)
= (x-y)(x+y-2)
b,2x+2y-x²-xy
= 2(x+y) - x(x+y)
= (x+y)(2-x)
c,3a²-6ab+3b²-12c²
= 3(a² - 2ab + b² - 4c²)
= 3[(a-b)² - 4c²)
= 3(a-b-2c)(a-b+2c)
d,x²-25+y²+2xy
= (x+y)² - 25
= (x+y+5)(x+y-5)

e) a²+2ab+b²-ac-bc

= (a+b)²-c(a+b)

= (a+b)( a+b-c)

f) x²-2x-4x²-4y

= -3x²-2x-4y

= -(3x²+2x+4y)

g)x²y-x³-9y+9x

= x²(y-x)-9(y-x)

= (y-x)(x²-9)

h) x²(x-1)+16(1-x)

= x²(x-1)-16(x-1)

= (x-1)(x²-16)

= (x-1)(x-4)(x+4)

n) 81x²-6yz-9y²-z²

= (9x)²-[(3y)²+6yz+z²]

=(9x)²-(3y+z)²

=(9x+3y+z)(9x-3y-z)

m) xz- yz-x²+2xy-y²

= z(x-y)-(x²-2xy+y²)

= z(x-y)-(x-y)²

= (x-y)(z-x+y)

 p) x² + 8x + 15

= x² + 3x + 5x + 15

= x(x+3) + 5(x+3)

= (x+3)(x+5)

k) x² - x - 12

= x² + 3x - 4x - 12

= x(x+3) - 4(x+3)

= (x+3)(x-4)

\(Q=\dfrac{2-\dfrac{c}{a}-\dfrac{2b}{a}+\left(\dfrac{b}{a}\right)\left(\dfrac{c}{a}\right)}{1-\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{2-mn+2\left(m+n\right)-mn\left(m+n\right)}{1+m+n+mn}\)

\(Q=\dfrac{\left(2-mn\right)\left(m+n+1\right)}{\left(m+1\right)\left(n+1\right)}\ge\dfrac{\left[8-\left(m+n\right)^2\right]\left(m+n+1\right)}{\left(m+n+2\right)^2}\)

Đặt \(m+n=t\Rightarrow0\le t\le2\)

\(Q\ge\dfrac{\left(8-t^2\right)\left(t+1\right)}{\left(t+2\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{\left(2-t\right)\left(4t^2+15t+10\right)}{4\left(t+2\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(t=2\) hay \(m=n=1\)