Biến x đâu bạn ?

-1/5

nha

>: kệ đề bài nhá, bài cho j lm cái đó.

a, \(\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)

\(\frac{-5}{7}+\frac{3}{4}-\frac{1}{5}-\frac{2}{7}+\frac{1}{4}=-1+1-\frac{1}{5}=-\frac{1}{5}\)

b, \(\frac{-3}{31}+\frac{-6}{17}+\frac{1}{25}+\frac{-28}{31}+\frac{-11}{17}+\frac{-1}{5}\)

\(=\frac{-3}{31}-\frac{6}{17}+\frac{1}{25}-\frac{28}{31}-\frac{11}{17}-\frac{1}{5}=-1-1-\frac{4}{25}==-2-\frac{4}{25}=-\frac{54}{25}\)

Sai j sửa hộ.

a) \(P=\frac{3x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(P=\frac{3\left(x-9\right)}{\left(x-3\right)\left(x-2\right)}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

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

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

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

\(P=\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}\) (*)

b) Thay \(x=-\frac{1}{2}\) vào (*) ta có:

\(P=\frac{2-\left(-\frac{1}{2}\right)^2}{\left[\left(-\frac{1}{2}\right)-2\right]\left[3-\left(-\frac{1}{2}\right)\right]}=\frac{2-\frac{1}{4}}{-\frac{5}{2}.\frac{7}{2}}=-\frac{\frac{7}{4}}{\frac{5}{2}.\frac{7}{2}}=-\frac{7}{35}=-\frac{1}{5}\)

c) \(\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}< 0\)

\(\Leftrightarrow2-x^2< 0\)

\(\Leftrightarrow-x^2< -2\)

\(\Leftrightarrow x^2>2\)

\(\Leftrightarrow\hept{\begin{cases}x< -\sqrt{2}\\-\sqrt{2}< x< \sqrt{2}\\x>2\end{cases}}\)

Vậy: ...

a) Xem lại đề

b) Ta có: \(2x=4y=5z\)=> \(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{5}}\) => \(\frac{2x}{1}=\frac{3y}{\frac{3}{4}}=\frac{z}{\frac{1}{5}}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{2x}{1}=\frac{3y}{\frac{3}{4}}=\frac{z}{\frac{1}{5}}=\frac{2x-3y-z}{1-\frac{3}{4}-\frac{1}{5}}=\frac{1}{\frac{1}{20}}=20\)

=> \(\hept{\begin{cases}\frac{x}{\frac{1}{2}}=20\\\frac{y}{\frac{1}{4}}=20\\\frac{z}{\frac{1}{5}}=20\end{cases}}\) => \(\hept{\begin{cases}x=20.\frac{1}{2}=10\\y=20.\frac{1}{4}=5\\z=20.\frac{1}{5}=4\end{cases}}\)

Vậy x = 10; y = 5 và z = 4

a)\(\frac{x}{5}=\frac{y}{6};\frac{y}{2}=\frac{z}{3}\)va \(x^3-2x^2y+z^3\)

\(\frac{11}{8}+\frac{13}{6}=\frac{85}{x}\)

\(\frac{85}{24}=\frac{85}{x}\)

\(x=24\)

\(2x-\frac{2}{11}=1\frac{1}{5}\)

\(2x-\frac{2}{11}=\frac{6}{5}\)

\(2x=\frac{76}{55}\)

\(x=\frac{38}{55}\)

a. \(\frac{11}{8}\)+ \(\frac{13}{6}\)= \(\frac{85}{x}\)

                    \(\frac{85}{24}\) =\(\frac{85}{x}\)

                             x   = \(\frac{24\cdot85}{85}\)

                             x   = 24

b. 2x - \(\frac{2}{11}\)=1\(\frac{1}{5}\)

     2x- \(\frac{2}{11}\)= \(\frac{6}{5}\)

     2x             = \(\frac{6}{5}\)+\(\frac{2}{11}\)

       x             = \(\frac{76}{55}\): 2

       x             = \(\frac{38}{55}\)

Học tốt nha ~~~~~ ỌvỌ

e, \(\frac{11}{8}+\frac{13}{6}=\frac{85}{x}\)

\(\Leftrightarrow\frac{85}{24}=\frac{85}{x}\)

\(\Leftrightarrow x=24\)

Vậy.......................

i dont no ok

Ta có :\(\frac{3}{x}+\frac{4}{y}+\frac{5}{z}=6\)

\(\Leftrightarrow\frac{6}{2x}+\frac{12}{3y}+\frac{20}{4z}=6\)

\(\Leftrightarrow\frac{6}{2x}+\frac{12}{2x}+\frac{20}{2x}=6\)

\(\Leftrightarrow\frac{6+12+20}{2x}=6\)

\(\Leftrightarrow\frac{19}{x}=6\)

\(\Leftrightarrow x=\frac{19}{6}\)

\(\Leftrightarrow\frac{2}{3}x=\frac{2}{3}.\frac{19}{6}=\frac{19}{9}=y\)

\(\Leftrightarrow\frac{3}{4}y=\frac{3}{4}.\frac{19}{9}=\frac{19}{12}=z\)

Vậy \(\hept{\begin{cases}x=\frac{19}{6}\\y=\frac{19}{9}\\z=\frac{19}{12}\end{cases}}\)