Ta có:\(A=\dfrac{xy}{x+y}+\dfrac{yz}{y+z}+\dfrac{zx}{z+x}\)

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

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

Ta có:\(\dfrac{x^2}{x+y}+\dfrac{x+y}{9}\ge2\sqrt{\dfrac{x^2}{x+y}.\dfrac{x+y}{9}}=\dfrac{2x}{3}\)

Tương tự,ta có:\(\dfrac{y^2}{y+z}+\dfrac{y+z}{9}\ge\dfrac{2y}{3};\dfrac{z^2}{z+x}+\dfrac{z+x}{9}\ge\dfrac{2z}{3}\)

Cộng vế với vế ta có:

\(\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}+\dfrac{2\left(x+y+z\right)}{4}\ge\dfrac{2\left(x+y+z\right)}{3}\)

\(\Leftrightarrow\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}\ge\dfrac{2\left(x+y+z\right)}{3}-\dfrac{2\left(x+y+z\right)}{4}=\dfrac{2.9}{3}-\dfrac{9}{2}=\dfrac{3}{2}\)

\(\Rightarrow A\le9-\dfrac{3}{2}=\dfrac{15}{2}\)

Dấu "=" xảy ra ⇔ x=y=z=3

Vậy,Max A=\(\dfrac{15}{2}\) ⇔ x=y=z=3

nếu khó nhìn để mik đánh lại

a)Ta có:2x⁴-2x³+x²+x+a

     = 2x³(x-2)+2x²(x-2)+5x(x-2)+11(x-2)+a+22

     = (x-2)(2x³+2x²-5x+11)+(a+22)

Để (x-2)(2x³+2x²-5x+11)+(a+22)⋮(x-2) thì a+22=0⇔a=-22

b)Ta có:2x³-3x²+x+a

         = 2x²(x+2)-5x(x+2)+11(x+2)+(a-22)

        = (x+2)(2x²-5x+11)+(a-22)

Để (x+2)(2x²-5x+11)+(a-22)⋮(x+2) thì a-22=0⇔a=22

a)R={75;76;77;...;84;85}

   S={75;76;77;...;90;91}

b)Tập hợp R có số phần tử là:(85-75):1+1=11 (phần tử)

   Tập hợp S có số phần tử là:(91-75):1+1=17 (phần tử)

C.vòng cung.

1.m_NaCl=200.10%+300.20%=80(g)

\(C\%=\dfrac{80}{200+300}.100\%=16\%\)

\(m_{H_2O}=272-72=200\left(g\right)\)

\(S_{NaCl\left(25^oC\right)}=\dfrac{72}{200}.100=36\left(g\right)\)

\(m_{H_2O}=272-72=200\left(g\right)\)

\(S_{NaCl\left(250^oC\right)}=\dfrac{72}{200}.100\%=36\%\)