Sang h.vn mà hỏi đây là trang hỏi t,v,a mà (chủ yếu là toán)

ĐKXĐ: x;y>=4

\(2.\left(x\sqrt{y-4}+y\sqrt{x-4}\right)=xy\)

\(\Leftrightarrow x.\sqrt{4}.\sqrt{y-4}+y.\sqrt{4}.\sqrt{x-4}=xy\)

Theo AM-GM ta có:

\(VT\le x.\frac{y}{2}+y.\frac{x}{2}=xy=VP\)

=> VT=VP<=> x=y=8

Vậy x=y=8

Tự đặt đ/k

Ko viết lại đề

\(\Leftrightarrow x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\)=0

\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)

Tự giải tiếp nha :v, đúng thì tk nha :3

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

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

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

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

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

\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}}{x+2\sqrt{x}}\)

\(A=\frac{4}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x+2}\right)}\)

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

\(A=\frac{4}{\left(\sqrt{x}+2\right)^3\left(\sqrt{x}-2\right)}\)