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

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

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

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

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

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

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

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

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

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

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

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

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

haha em không biết câu trả lời em mới học lớp 6

A B C D I J O K

a) Gọi tiếp điểm của \(\left(I\right),\left(J\right)\) là \(K\)

Ta có \(\frac{DA+DB-AB}{2}=DK=\frac{DA+DC-AC}{2}\Leftrightarrow AB-AC=DB-DC\)

Vậy điểm \(D\) nằm trên cạnh \(BC\) và thỏa \(AB-AC=DB-DC\).

Từ đó, ta dựng điểm \(D\) như sau: (Giả sử \(AB>AC\))

B1: Lấy \(E\) trên cạnh \(AB\) sao cho \(AE=AC\)

B2: Lấy \(F\) trên cạnh \(BC\) sao cho \(BF=BE\)

B3: Lấy trung điểm \(D\) của \(CF\)

b) Dễ thấy:

\(\widehat{OAC}=\widehat{OAJ}+\widehat{JAC}=90^0-\widehat{AIJ}+90^0-\widehat{AJI}=\widehat{IAJ}\)

Tương tự \(\widehat{OAB}=\widehat{IAJ}\). Vậy \(O\) nằm trên phân giác của \(\widehat{BAC}.\)

\(a,A=\frac{9}{\sqrt{x}+4}< \frac{3}{2}\)

\(\frac{9}{\sqrt{x}+4}-\frac{3}{2}< 0\)

\(\frac{18-3\sqrt{x}-12}{2\sqrt{x}+8}< 0\)

\(\frac{6-3\sqrt{x}}{2\sqrt{x}+8}< 0\)

\(2\sqrt{x}+8>0< =>6-3\sqrt{x}< 0\)

\(3\sqrt{x}>6\)

\(x>4\)

\(A=\frac{9}{\sqrt{x}+4}=2\sqrt{x}-9\)

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

\(2x+8\sqrt{x}-9\sqrt{x}-36=9\)

\(2x-\sqrt{x}-45=0\)

giải pt bậc 2 là ra nha bạn

\(a,C=\frac{7}{\sqrt{x}+3}< 1\)

\(C=\frac{7}{\sqrt{x}+3}-1< 0\)

\(C=\frac{7-\sqrt{x}-3}{\sqrt{x}+3}< 0\)

\(C=\frac{4-\sqrt{x}}{\sqrt{x}+3}< 0\)

\(\sqrt{x}+3>0< =>4-\sqrt{x}< 0\)

\(\sqrt{x}>4\)

\(x>16\)

\(b,\sqrt{x}+2C=\sqrt{x}+\frac{14}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\frac{14}{\sqrt{x}+3}-3\)

\(\sqrt{x}+3+\frac{14}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{14}{\sqrt{x}+3}}=2\sqrt{14}\)

\(\sqrt{x}+2C\ge2\sqrt{14}-3\)dấu "=" xảy ra khi \(\sqrt{x}+3=\frac{14}{\sqrt{x}+3}\)

\(x+6\sqrt{x}+9=14\)

\(x+6\sqrt{x}-5=0\)

rồi bạn giải pt bậc 2 

\(< =>MIN=2\sqrt{14}-3\)

\(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)=4\sqrt{18}-6+2\sqrt{3}\)

\(=12\sqrt{2}-6+2\sqrt{3}\)