a) Ta có: \(A=\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)

\(=\sqrt{1+2\cdot1\cdot\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}\)

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

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

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

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

\(=\frac{2\sqrt{2}+2}{1+\sqrt{2}}\)

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

b) Ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right)\cdot\frac{\sqrt{x}+3}{x+9}\)

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

\(=\frac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{1}{\sqrt{x}-3}\)

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

\(=\frac{1}{\sqrt{x}-3}\)(đpcm)

Trả lời nhanh giúp mình với mình cần gấp lắm

Mình giải câu a thấy số xấu và câu b không thỏa dạng nên mình sửa đề lại nha. Hi vọng đúng với đề gốc của bạn.

a. Thay x=25 vào A ta được: A=\(\frac{7}{\sqrt{25}+8}=\frac{7}{13}\)

b. B=\(\frac{\sqrt{x}}{\sqrt{x-3}}+\frac{2\sqrt{x}-24}{x-9}\)

\(\Leftrightarrow B=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\frac{2\sqrt{x}-24}{x-9}\)

\(\Leftrightarrow B=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow B=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)(\sqrt{x}+3)}\)

\(\Leftrightarrow B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow B=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)

c. ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne-8\\x\ne-3\end{matrix}\right.\)

P=A.B=\(\frac{7}{\sqrt{x}+8}.\frac{\sqrt{x}+8}{\sqrt{x}+3}=\frac{7}{\sqrt{x}+3}\)

Để P nguyên thì \(\sqrt{x}+3\:\in\) Ư(7) \(\Leftrightarrow\sqrt{x}+3\in\){\(\pm1;\pm7\)}

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

\(\sqrt{x}+3=-1\text{​​}\Leftrightarrow\sqrt{x}=-4\) (KTM)

\(\sqrt{x}+3=7\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\) (TM)

\(\sqrt{x}+3=-7\Leftrightarrow\sqrt{x}=-10\:\left(KTM\right)\)

Vậy...

Khi x=25

=> A=\(\frac{7}{\sqrt{25+8}}=\frac{7}{\sqrt{\text{3}\text{3}}}\)=\(\frac{7\sqrt{33}}{33}\)

b)  B= \(\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}+\frac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

B=  \(\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\)

B=  \(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)

a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

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

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

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

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

\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)

b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)

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

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

sao ko ai làm hộ tôi vậy 

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