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

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

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

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

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

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

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

\(\Leftrightarrow x^2+2x+1+x^4+1+x^2+2x^2-2x-2x^3=2\left(x^3+1\right)\)

\(\Leftrightarrow4x^2+2+x^4-2x^3=2x^3+2\)

\(\Leftrightarrow x^4-2x^3-2x^3+4x^2=2-2\)

\(\Leftrightarrow x^4-4x^3+4x^2=0\)

\(\Leftrightarrow x^2\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-4x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-2\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Bạn tự tìm ĐKXĐ rồi so sánh kết quả nhé!

\(\sqrt{a}+2\sqrt{a}-1+\sqrt{a}-2\sqrt{a}-1\)

\(=2\sqrt{a}-2\)

vậy thôi à??

ko phải 

\(\frac{2}{\sqrt{7}-\sqrt{5}}+\frac{6}{\sqrt{11}+\sqrt{5}}\)

\(=\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}+\frac{6\left(\sqrt{11}-\sqrt{5}\right)}{\left(\sqrt{11}+\sqrt{5}\right)\left(\sqrt{11}-\sqrt{5}\right)}\)

\(=\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{7-5}+\frac{6\left(\sqrt{11}-\sqrt{5}\right)}{11-5}\)

\(=\frac{2\left(\sqrt{7}+\sqrt{5}\right)}{2}+\frac{6\left(\sqrt{11}-\sqrt{5}\right)}{6}\)

\(=\sqrt{7}+\sqrt{5}+\sqrt{11}-\sqrt{5}\)

\(=\sqrt{7}+\sqrt{11}\)

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

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

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

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

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

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

\(=\sqrt{3}\)

\(B=\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)

\(=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}+1\right)}\)

\(=\frac{\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)}+\frac{\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)}\)

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

\(=\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=\frac{12}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=\frac{12}{5-1}\)

\(=\frac{12}{4}\)

\(=3\)

Bn tự vẽ hình nhé...

a)

AB⊥CD (GT) => CIB =90 độ (1)

AEB=90độ ( góc nt chắn nữa dg tròn) (2)

Từ (1)và(2) tứ giác BEFI nội tiếp

b)

Xét ΔAFC và Δ ACE có

  A( góc chung)

  C=E( vì 2 góc cùng chắn 2 cung AC và AD bằng nhau)

=>ΔAFC∼Δ ACE

=> AC/AE=AF/AC

=> AE.AF=AC²