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

\(\Rightarrow2-x^2+2-\frac{1}{x^2}+2\sqrt{\left(2-x^2\right)\left(2-\frac{1}{x^2}\right)}=16-8\left(x+\frac{1}{x}\right)+\left(x+\frac{1}{x}\right)^2\)

\(\Rightarrow4-\left(x^2+\frac{1}{x^2}\right)+2\sqrt{5-2\left(x^2+\frac{1}{x^2}\right)}=16-8\left(x+\frac{1}{x}\right)+\left(x+\frac{1}{x}\right)^2\)

\(\Rightarrow x^2+\frac{1}{x^2}+2\sqrt{5-2\left(x^2+\frac{1}{x^2}\right)}=8\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)^2-12\)

Đặt \(a=x+\frac{1}{x}\Rightarrow\left|a\right|=\left|x+\frac{1}{x}\right|=\left|x\right|+\frac{1}{\left|x\right|}\ge2\Rightarrow\left|a\right|\ge2\)

Phươn trình trở thành:

\(a^2-2+2\sqrt{5-2\left(a^2-2\right)}=8a-a^2-12\)

Tớ nghĩ là theo cách này có vẻ khả quan

Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)

\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(A=\frac{11}{3}\)(1)

+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)

\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)

=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)

=> \(\frac{11}{3}-x=1\)

=> \(x=\frac{11}{3}-1=\frac{8}{3}\)

Vậy x = 8/3

\(bn\)\(xem\)\(lai\)\(giup\)\(mk\)\(cho\)\(\frac{x+522}{7}\)\(neu\)\(thay\)\(bang\)\(\frac{x+552}{7}\)\(thi\)\(dug\)\(hon\)

thế thì bạn giải thử xem cô t ra đề thế mà ừ thì cứ cho là x + 552 cx đc

a) \(\frac{1}{3}-\left(\frac{1}{2}+\frac{1}{8}\right)\)

=   \(\frac{1}{3}-\left(\frac{4}{8}+\frac{1}{8}\right)\)

=     \(\frac{1}{3}-\frac{5}{8}\)

= \(\frac{8}{24}-\frac{15}{24}\)

= \(\frac{-7}{24}\)

b) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{13}+\frac{1}{8}\)

= \(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)\)+ \(\frac{1}{13}\)

= \(\left(\frac{4}{8}-\frac{2}{8}+\frac{1}{8}\right)+\frac{1}{13}\)

=                 \(\frac{1}{8}+\frac{1}{13}\)

=                 \(\frac{13}{104}+\frac{8}{104}\)

=                        \(\frac{23}{104}\)

c) \(13\frac{2}{7}:\left(\frac{-8}{9}\right)+2\frac{5}{7}:\left(\frac{-8}{9}\right)\)

= \(\left(13\frac{2}{7}+2\frac{5}{7}\right):\left(\frac{-8}{9}\right)\)

=         \(16:\left(\frac{-8}{9}\right)\)

=         -18

Câu 2 sai đề, thử rồi

Từ đề bài trên, ta có:

\(=\left(\frac{-5}{13}+\frac{-8}{13}\right)+\left(\frac{1}{4}+\frac{3}{4}\right)+\frac{2}{11}=-1+1+\frac{2}{11}=\frac{2}{11}\)

P/s: Sai đề bài bạn nhé. 3/4 chứ không phải là 3/14 đâu :3

Học tốt!