bbgfhfygfdsdty64562gdfhgvfhgfhhhhh

\hvhhhggybhbghhguyg

a: \(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{y}+\sqrt{x}}=\frac{x+y}{\sqrt{x}+\sqrt{y}}\)

Ta có: \(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{\left(x+y\right)_{}\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)

\(\) \(=\frac{x^2-\sqrt{xy}\left(x+y\right)-y^2-x^2+y^2}{\sqrt{xy}\left(x-y\right)}=\frac{-\left(x+y\right)}{x-y}\)

b: Thay x=3; \(y=4+2\sqrt3\) vào A, ta được:

\(A=\frac{-\left(3+4+2\sqrt3\right)}{3-\left(4+2\sqrt3\right)}=\frac{-7-2\sqrt3}{-2\sqrt3-1}=\frac{7+2\sqrt3}{2\sqrt3+1}\)

\(=\frac{\left(7+2\sqrt3\right)\left(2\sqrt3-1\right)}{12-1}=\frac{14\sqrt3-7+12-2\sqrt3}{11}=\frac{12\sqrt3+5}{11}\)

Đề a,b bạn ghi mik ko hiểu

c)Ta có : \(x+y=a=>x^2+y^2+2xy=a^2\)

Mà  \(x^2+y^2=b\)nên\(b+2xy=a^2=>xy=\frac{a^2-b}{2}\)

\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\)

Thay \(x+y=a\) ; \(x^2+y^2=b\)và \(xy=\frac{a^2-b}{2}\)ta có : \(x^3+y^3=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)

\(x+y=1\Rightarrow x=1-y\)        

\(A=x^3+y^3+xy\)

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

\(=x^2+y^2\) (vì x + y = 1)

\(=\left(1-y\right)^2+y^2\)

\(=2y^2-2y+1\)

\(=2\left(y^2-y+\frac{1}{4}\right)+\frac{1}{2}=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall y\)

Dấu "=" xảy ra khi: \(y-\frac{1}{2}=0\Rightarrow y=\frac{1}{2}\Rightarrow x=1-y=\frac{1}{2}\)

Vậy GTNN của A là \(\frac{1}{2}\)khi \(x=y=\frac{1}{2}\)

\(A=x^3+y^3+xy=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(=x^2-xy+y^2+xy=x^2+y^2\ge\frac{\left(x+y\right)^2}{2}=\frac{1}{2}\)

Nên min A là \(\frac{1}{2}\) khi \(x=y=\frac{1}{2}\)

(x+y)(x²-xy+y²)+(x-y)(x²+xy+y²)

=x³-x²y+xy²+x²y-xy²+y³+x³+x²y+xy²-x²y-xy²-y³

=2x³

Thay x=3 ta có:

2x³=2 x 3³=2x27=54

cảm ơn bạn