Với \(a>0,b>0\) ta luôn có \(a+b\ge2\sqrt{ab}\)

M = \(\dfrac{x^2+y^2}{xy}=\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{3x}{y}+\left(\dfrac{x}{4y}+\dfrac{y}{x}\right)\)

Ta có: \(\left(\dfrac{x}{4y}+\dfrac{y}{x}\right)\ge2\sqrt{\dfrac{x}{4y}\cdot\dfrac{y}{x}}=1\)

Mặt khác: \(x\ge2y\) \(\Rightarrow\dfrac{3x}{4y}\ge\dfrac{3}{2}\)

Do đó \(M\ge\dfrac{5}{2}\) . Dâu ''='' xảy ra khi \(x=2y\)

Vậy giá trị nhỏ nhất của M là \(\dfrac{5}{2}\) \(\Leftrightarrow x=2y\)

\(\dfrac{x^2+y^2}{xy}\) ≥ \(\dfrac{2\sqrt{x^2y^2}}{xy}\) ≥ \(\dfrac{2xy}{xy}\) ≥ 2

Các bn check hộ mk nhé...

\(\left(x+y-1\right)\left(x+y+1\right)=x^2+xy-x+xy+y^2-y+x+y-1\\ =x^2+\left(xy+xy\right)+\left(-x+x\right)+y^2+\left(-y+y\right)-1\\ =x^2+2xy+y^2-1\\ =>B\)

a.

(x^2 + y^2 - 2xy) + (x^2 + y^2 + 2xy)

= x^2 + y^2 - 2xy + x^2 + y^2 + 2xy

= (x^2 + x^2) + (y^2 + y^2) + (2xy - 2xy)

= 2x^2 + 2y^2

b.

(x^2 + y^2 - 2xy) - (x^2 + y^2 + 2xy)

= x^2 + y^2 - 2xy - x^2 - y^2 - 2xy

= (x^2 - x^2) + (y^2 - y^2) - (2xy + 2xy)

= -4xy

phân tích đa thức có dạng m² + n ( n thuộc z)

bàn làm giúp mình đk ko ạ!

a) \(\dfrac{2x^2-2xy}{x^2+x-xy-y}\) \(\left(x\ne y;x\ne-1\right)\)

\(=\dfrac{2x\left(x-y\right)}{x\left(x+1\right)-y\left(x+1\right)}\)

\(=\dfrac{2x\left(x-y\right)}{\left(x-y\right)\left(x+1\right)}\)

\(=\dfrac{2x}{x+1}\)

b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

\(=\dfrac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)

\(=\dfrac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x-y+z\right)\left(x+y+z\right)}\)

\(=\dfrac{x+y-z}{x-y+z}\)

a: \(\dfrac{x^2+2xy+y^2}{x+y}=x+y\)

b: \(\dfrac{64x^3+1}{4x+1}=16x^2-4x+1\)