Câu 1: 

Thay x=0 vào y=x+1, ta được:

y=0+1=1

Thay y=0 vào y=x+1, ta được:

x+1=0

hay x=-1

vậy: A(-1;0); B(0;1)

\(AB=\sqrt{\left(-1-0\right)^2+\left(0-1\right)^2}=\sqrt{2}\)

\(C_{OAB}=OA+OB+AB=2+\sqrt{2}\)

\(S_{OAB}=\dfrac{OA\cdot OB}{2}=\dfrac{1}{2}\)

a: \(x=2\sqrt{2}+2-2\sqrt{2}+2=4\)

Thay x=4 vào B, ta được:

\(B=\dfrac{2-3}{2+1}=\dfrac{-1}{3}\)

\(K=\dfrac{2\sqrt{3a+1}+2\sqrt{3b+1}+2\sqrt{3c+1}}{2}\)\(\le\)\(\dfrac{3a+1+4+3b+1+4+3c+1+4}{4}=\dfrac{24}{4}=6\)

Vậy \(K_{max}=6\)

Dấu bằng xảy ra khi a=b=c=1

1, Với x > 0 ; x khác 4 

\(A=\left(\dfrac{\sqrt{x}-1}{x-4}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}\right):\dfrac{x\sqrt{x}}{\left(x-4\right)^2}\)

\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right):\dfrac{x\sqrt{x}}{\left(x-4\right)^2}\)

\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(x-4\right)\left(\sqrt{x}+2\right)}:\dfrac{x\sqrt{x}}{\left(x-4\right)^2}\)

\(=\dfrac{2\sqrt{x}\left(x-4\right)^2}{x\sqrt{x}\left(x-4\right)\left(\sqrt{x}+2\right)}=\dfrac{2\left(\sqrt{x}-2\right)}{x}\)

2, Ta có \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Thay vào ta được \(A=\dfrac{2\left(\sqrt{3}-1\right)}{4+2\sqrt{3}}=-5+3\sqrt{3}\)

3, Ta có \(\dfrac{2\left(\sqrt{x}-2\right)}{x}-\dfrac{1}{4}\ge0\Leftrightarrow\dfrac{8\left(\sqrt{x}-2\right)-x}{4x}\ge0\)

\(\Rightarrow-x+8\sqrt{x}-16\ge0\Leftrightarrow-\left(\sqrt{x}-4\right)^2\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-4\right)^2\le0\Leftrightarrow\sqrt{x}-4\le0\Leftrightarrow x\le16\)

Kết hợp đk vậy 0 < x =< 16 ; x khác 4 

\(tan30=\dfrac{AB}{AC}\)

\(AB=5.tan30\)

\(D\)

1) \(\dfrac{4\sqrt{x}-x-4}{x-4}\)

\(=\dfrac{-x+4\sqrt{x}-4}{x-4}\)

\(=\dfrac{-\left(x-4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{-\sqrt{x}+2}{\sqrt{x}+2}\)

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

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

\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

3) \(\dfrac{x-9}{x\sqrt{x}-27}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}\right)^3-3^3}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)}\)

\(=\dfrac{\sqrt{x}+3}{x+3\sqrt{x}+9}\)

\(1,\dfrac{4\sqrt{x}-x-4}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{-\left(x-4\sqrt{x}+4\right)}{\sqrt{x^2}-2^2}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}=\dfrac{2-\sqrt{x}}{\sqrt{x}+2}\)

\(2,\dfrac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}\left(dk:x,y\ge0\right)\\ =\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

\(3,\dfrac{x-9}{x\sqrt{x}-27}\left(dk:x\ge0\right)\\ =\dfrac{\sqrt{x^2}-3^2}{\sqrt{x^3}-3^3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(x+3\sqrt{x}+9\right)}=\dfrac{\sqrt{x}+3}{x+3\sqrt{x}+9}\)

Bạn muốn làm gì với phương trình này nhỉ? Nếu chỉ có điều kiện $x,y$ không âm và pt như thế này thì không tìm được giá trị $x,y$ cụ thể.

c: Ta có: \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)

\(=2-\sqrt{3}+2+\sqrt{3}\)

=4

d: Ta có: \(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

\(=\dfrac{a-2\sqrt{ab}+b+a+2\sqrt{ab}+b}{a-b}\)

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

\(A=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\\ A=\sqrt{x}+1-\sqrt{x}+2=3\)

Anh có thể giải từng bước giúp e ko ạ e ch hỉu lắm í ạ em cảm ơn