đây là câu hỏi mà bạn mình nhờ gửi

\(\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}+1}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)

\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2\sqrt{x^3}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{12x+22\sqrt{x}+2\sqrt{x^3}+12}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)

\(=\frac{2\left(6x+11\sqrt{x}+\sqrt{x^3}+6\right)}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)

\(=2\) (ko phụ thuộc vào biến ) (đpcm)

\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(\sqrt{3}+2\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{\sqrt{5}+2}+\sqrt{x}}\\ =\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+1-\sqrt{x}=1\)

sao ko ai làm hộ tôi vậy 

\(R=\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)

\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

=\(\frac{2x\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x\sqrt{x}+6x+5x+10\sqrt{x}+x+\sqrt{x}+10\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x\sqrt{x}+12x+21\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

@@@@@@@@@@@ Đề sai hay mình sai??@@@@@@@@@@

Ta có \(\sqrt[3]{2-\sqrt{3}}\times\sqrt[6]{7+4\sqrt{3}}\)

= \(\sqrt[3]{2-\sqrt{3}}\times\sqrt[3]{2+\sqrt{3}}\)

= 1

Và \(\sqrt[4]{9-4\sqrt{5}}\times\sqrt{2+\sqrt{5}}\)

= \(\sqrt{\sqrt{5}-2}\times\sqrt{2+\sqrt{5}}\)

= 1

Vậy A = \(\frac{\sqrt{x}+1-x}{1+\sqrt{x}}\)\(=1-\frac{x}{1+\sqrt{x}}\)

Vậy A phải phụ thuộc vào x. Có thể đề sai

\(K=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)

\(=\sqrt{x}+\dfrac{1-x}{1+\sqrt{x}}=\sqrt{x}+1-\sqrt{x}=1\)

Vậy K không phụ thuộc vào x

\(=\dfrac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x\sqrt{x}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+11}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x\sqrt{x}+12x+22\sqrt{x}+13}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

biểu thức này có phụ thuộc vào biến nha bạn