a/ Bạn tự tìm ĐKXĐ

\(A=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)

Xét 

\(=\frac{\sqrt{x}-x\sqrt{y}+1-\sqrt{xy}+xy+\sqrt{xy}+x\sqrt{y}+\sqrt{x}+1-xy}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)

\(=\frac{2\sqrt{x}+2}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)

\(=\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)

\(=\frac{xy-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)

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

\(\Rightarrow A=\frac{2\left(\sqrt{x}+1\right)}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}:\frac{2\sqrt{xy}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}=\frac{1}{\sqrt{xy}}\)

b/ Áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) với \(a=\frac{1}{\sqrt{x}},b=\frac{1}{\sqrt{y}}\) được : 

\(A=\frac{1}{\sqrt{x}.\sqrt{y}}\le\frac{1}{4}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2=\frac{1}{4}.6^2=9\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\end{cases}}\Leftrightarrow x=y=\frac{1}{9}\)

Vậy ........................................................

\(6=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\ge\frac{2}{\sqrt[4]{xy}}\)\(\Leftrightarrow\)\(\frac{1}{\sqrt{xy}}\le9\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{9}\)

1, A= y^3(1-y)^2 = 4/9 . y^3 . 9/4 (1-y)^2

= 4/9 .y.y.y . (3/2-3/2.y)^2

=4/9 .y.y.y (3/2-3/2.y)(3/2-3/2.y)

<= 4/9 (y+y+y+3/2-3/2.y+3/2-3/2.y)^5

=4/9 . 243/3125

=108/3125

Đến đó tự giải

Thử sức với bài 1 xem thế nào :vv
x>0 => 0<x<=1 
f(x)=x^2(1-x)^3
Xét f'(x) = -(x-1)^2x(5x-2) 
Xét f'(x)=0 -> nhận x=2/5 và x=1thỏa mãn đk trên .
 Thử x=1 và x=2/5 nhận x=2/5 hàm số Max tại ddk 0<x<=1 (vậy x=1 loại)
P/s: HS cấp II hong nên làm cách này nhé em :vv 
 

tích mình với

ai tích mình

mình tích lại

thanks

Tích mình đi mình tích lại

1.

a,

\(A\text{ xác định }\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne0\end{matrix}\right.\)

\(\text{Vậy A xác định }\Leftrightarrow x>0\text{ và }x\ne1\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)

\(=\left(\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}-1\right)\)

\(=\frac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\frac{x-2}{\sqrt{x}}\)

b, \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)

\(A=\frac{x-2}{\sqrt{x}}=\frac{3-2\sqrt{2}-2}{\sqrt{2}-1}\)

\(=\frac{1-2\sqrt{2}}{\sqrt{2}-1}=-\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}+1\right)}{\sqrt{2}-1}=-3-\sqrt{2}\)

1.a)\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne0;x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)

\(=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)

\(=\frac{x+2}{\sqrt{x}}\)

b)\(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)

Thay x vào A ta được

\(A=\frac{3-2\sqrt{2}+2}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

\(=\frac{\sqrt{2}+6-1-3\sqrt{2}}{\sqrt{2}-1}\)(vì \(\sqrt{2}>1\))

\(=\frac{\sqrt{2}\left(1+3\sqrt{2}\right)-\left(1+3\sqrt{2}\right)}{\sqrt{2}-1}\)

\(=\frac{\left(\sqrt{2}-1\right)\left(1+3\sqrt{2}\right)}{\sqrt{2}-1}=1+3\sqrt{2}\)

c)\(A=\frac{x+2}{\sqrt{x}}=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}\)

Dấu "=" xảy ra khi \(x=2\)

\(\Rightarrow Min_A=2\sqrt{2}\) khi x=2