đề sai

\(a+\frac{1}{a}=b+\frac{1}{b}\Leftrightarrow a+\frac{1}{a}-b-\frac{1}{b}=0\Leftrightarrow\frac{a^2b+b-ab^2-a}{ab}=0\)\(\Leftrightarrow a^2b+b-ab^2-a=0\Leftrightarrow\left(a^2b-ab^2\right)-\left(a-b\right)=0\Leftrightarrow ab\left(a-b\right)-\left(a-b\right)=0\)\(\Leftrightarrow\left(a-b\right)\left(ab-1\right)=0\)

Do a,b khác nhau và khác o nên ab=1 => đpcm

Mình chọn giá trị biểu thức ban đàu là 2 nghe.  Do bạn đăng đề không rõ. Nếu là giá trị khác thì bạn tự thay số nghe.

\(P=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}=\left(\frac{b}{a+b}-1\right)+\left(\frac{c}{b+c}-1\right)+\left(\frac{a}{c+a}-1\right)+3\)\(=\frac{b-a-b}{a+b}+\frac{c-b-c}{b+c}+\frac{a-c-a}{c+a}+3=-\frac{a}{a+b}-\frac{b}{b+c}-\frac{c}{c+a}+3\)

\(=-\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)+3\)

Thay giá trị ban đầu đề bài cho \(\Rightarrow P=-2+3=1\) 

_Tao quên cách làm rồi chi ạ ( Bi = Jin :)) )

chán ghê

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\)

\(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

\(=\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\)

\(=y\left(\frac{1}{x}+\frac{1}{z}\right)+x\left(\frac{1}{z}+\frac{1}{y}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}\)

\(=-1-1-1=-3\)

P+3=\(\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)

P+3=\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0.\left(x+y+z\right)=0\)

=> P=\(-3\)

Chuc ban hoc tot

\(a,4\frac{1}{3}\left[\frac{1}{2}-\frac{1}{6}\right]\le x\le-\frac{2}{3}\left[\frac{1}{3}\cdot\frac{1}{2}-\frac{3}{4}\right]\)

=> \(\frac{13}{3}\left[\frac{3}{6}-\frac{1}{6}\right]\le x\le-\frac{2}{3}\left[\frac{1}{6}-\frac{3}{4}\right]\)

=> \(\frac{13}{3}\cdot\frac{1}{3}\le x\le-\frac{2}{3}\cdot\left[\frac{2}{12}-\frac{9}{12}\right]\)

=> \(\frac{13}{9}\le x\le-\frac{2}{3}\cdot\left[-\frac{7}{12}\right]\)

=> \(\frac{13}{9}\le x\le-\frac{1}{3}\cdot\left[-\frac{7}{6}\right]\)

=> \(\frac{13}{9}\le x\le\frac{7}{18}\)

Đến đây tự tìm x 

\(e,\frac{22}{15}-x=-\frac{8}{27}\)

=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)

=> \(x=\frac{22}{15}+\frac{8}{27}\)

=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)

\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)

=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)

=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)

=> \(\frac{2x-5}{5}=-\frac{5}{4}\)

=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)

=> \(2x=-\frac{5}{4}\)

=> \(x=-\frac{5}{8}\)

\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)

=> \(-\frac{9}{4}x+\frac{37}{4}=20\)

=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)

=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)

\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)

=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)

=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)

=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)

=> \(-10\le x\le-\frac{13}{7}\)

Đến đây tìm x