\(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

\(x+\sqrt{x+\frac{1}{4}+\sqrt{x+\frac{1}{4}}+\frac{1}{2}^2}=2\)

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

\(x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)

\(\sqrt{x+\frac{1}{4}}=\frac{3}{2}-x\)

\(x+\frac{1}{4}=\frac{9}{4}-3x+x^2\)

\(x^2-4x+2=0\)

\(\Delta=\left(-4\right)^2-\left(4.2.1\right)=8\)

\(\sqrt{\Delta}=2\sqrt{3}\)

\(\orbr{\begin{cases}x_1=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\left(TM\right)\\x_2=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\left(TM\right)\end{cases}}\)

B1.

\(\left(\sqrt{8}+\sqrt{72}-\sqrt{2}\right)\sqrt{2}\)

\(=\left(2\sqrt{2}+6\sqrt{2}-\sqrt{2}\right)\sqrt{2}\)

\(=14\)

\(3,\sqrt{10-2\sqrt{21}}.\left(\sqrt{7}+\sqrt{3}\right)\)

\(=\sqrt{\sqrt{3^2}-2\sqrt{3}\sqrt{7}+\sqrt{7^2}}.\left(\sqrt{7}+\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}.\left(\sqrt{7}+\sqrt{3}\right)\)

\(=-\left(\sqrt{3}-\sqrt{7}\right).\left(\sqrt{7}+\sqrt{3}\right)\)

\(=\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)\)

\(=7-3=4\)

\(4,\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)

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

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

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

B3

\(a,\frac{2}{3}\sqrt{9x+27}-\frac{3}{2}\sqrt{4x+12}-2=\sqrt{3+x}\)

\(Đkxđ:x\ge-3\)

\(\Leftrightarrow\frac{2}{3}\sqrt{9\left(x+3\right)}-\frac{3}{2}\sqrt{4\left(x+3\right)}-\sqrt{3+x}=2\)

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

\(\Leftrightarrow2\sqrt{x+3}-3\sqrt{x+3}-\sqrt{3+x}=2\)

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

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

\(\Leftrightarrow\left(\sqrt{x+3}\right)^2=\left(-1\right)^2\)    Bạn muốn mất căn thì phải bình phương lên nhé!
\(\Leftrightarrow x+3=1\)

\(\Leftrightarrow x=-2\)

Vậy x= -2

\(b,3\sqrt{x-1}+3\sqrt{\frac{4x-4}{9}}=\frac{2}{3}\sqrt{\frac{9x-9}{4}}+2\)

\(Đkxđ:x\ge1\)

\(\Leftrightarrow3\sqrt{x-1}+3\frac{\sqrt{4\left(x-1\right)}}{\sqrt{9}}=\frac{2}{3}.\frac{\sqrt{9\left(x-1\right)}}{\sqrt{4}}+2\)

\(\Leftrightarrow3\sqrt{x-1}+\frac{6}{3}\sqrt{x-1}=\frac{2}{3}.\frac{3}{2}\sqrt{x-1}+2\)

\(\Leftrightarrow3\sqrt{x-1}+\frac{6}{3}\sqrt{x-1}=\sqrt{x-1}+2\)

\(\Leftrightarrow3\sqrt{x-1}+\frac{6}{3}\sqrt{x-1}-\sqrt{x-1}=2\)

​​​\(\Leftrightarrow4\sqrt{x-1}=2\)​

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

\(\Leftrightarrow x-1=\frac{1}{4}\)

\(\Leftrightarrow x=\frac{5}{4}\)

Vậy \(x=\frac{5}{4}\)

Học tốt!

Xét (O) có A,B thuộc O

=> OA=OB mà OA=AB

=> OA=OB=AB

=> tam giác OAB đều

=> AOB= 60 độ

tính góc AOB ạ.

\(A=\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)

\(A=\frac{3\sqrt{2}+\sqrt{10}}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}-\frac{3\sqrt{2}-\sqrt{10}}{2\sqrt{5}+\sqrt{6-2\sqrt{5}}}\)

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

\(A=\frac{3\sqrt{2}+\sqrt{10}}{2\sqrt{5}+\sqrt{5}+1}-\frac{3\sqrt{2}-\sqrt{10}}{2\sqrt{5}+\sqrt{5}-1}\)

\(A=\frac{3\sqrt{2}+\sqrt{10}}{3\sqrt{5}+1}-\frac{3\sqrt{2}-\sqrt{10}}{3\sqrt{5}-1}\)

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

\(A=\frac{90+3\sqrt{50}-3\sqrt{2}-\sqrt{10}-90+3\sqrt{50}-3\sqrt{2}+\sqrt{10}}{44}\)

\(A=\frac{6\sqrt{50}-6\sqrt{2}}{44}=\frac{\sqrt{2}\left(6\sqrt{25}-6\right)}{44}=\frac{24\sqrt{2}}{44}=\frac{6\sqrt{2}}{11}\)

con cảm ơn nhiều ạ.

\(2\sqrt{x-1}-\sqrt{x+2}=2-x\left(ĐKXĐ:x\ge1\right)\)

\(\Leftrightarrow2\left(\sqrt{x-1}-1\right)-\left(\sqrt{x+2}-2\right)+\left(x-2\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\frac{2}{\sqrt{x-1}+1}-\frac{1}{\sqrt{x+2}+2}+1=0\left(2\right)\end{cases}}\)

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

Vì \(x\ge1\)\(\Rightarrow\frac{2}{\sqrt{x-1}+1}+\frac{\sqrt{x+2}+1}{\sqrt{x+2}+2}>0\)

=> (2) vô nghiệm )

Vậy nghiệm của pt là x= 2

Trả lời:

a, \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-3}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\) \(\left(đkxđ:x\ge0;x\ne9\right)\)

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

\(=\frac{x-3\sqrt{x}}{x-9}+\frac{2x+3\sqrt{x}-9}{x-9}-\frac{2x-\sqrt{x}-3}{x-9}\)

\(=\frac{x-3\sqrt{x}+2x+3\sqrt{x}-9-2x+\sqrt{x}+3}{x-9}\)

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