a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

Ta có : 

\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)

\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)

\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)

Nên \(x+50=0\)

\(\Rightarrow\)\(x=-50\)

Vậy \(x=-50\)

Chúc bạn học tốt ~ 

\(\frac{x-1}{21}=\frac{3}{x+1}\)

=> \(\left(x-1\right)\left(x+1\right)=21\cdot3\)

=> \(x^2-1=63\)

=> \(x^2=64\)

=> \(\orbr{\begin{cases}x^2=8^2\\x^2=\left(-8\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)

\(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)

=> \(\frac{12}{13}x=2\)

=> \(x=\frac{13}{6}\)

d, \(\frac{x-1}{21}=\frac{3}{x+1}\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=63\)

\(\Leftrightarrow x^2-1=63\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)

e, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)

\(\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)

a, 60%x + 2/3x =1/3.6 1/3

3/5x +2/3x =1/3.19/3

x.(3/5+2/3)=19/9

x.(9/15+10/15)=19/9

x.19/15=19/9

x=19/9:19/15

x=15/9 

Vậy x=15/9

b,3.(3x-1/2)^3 +1/9=0

3.(3x-1/2)^3= -1/9

(3x-1/2)^3= -1/9:3

(3x-1/2)^3= -1/27

(3x-1/2)^3=(-1/3)^3

3x-1/2= -1/3

3x= -1/3-1/2

3x= -2/6+(-3/6)

3x= -5/6

x= -5/6 :3

x=-5/18 

Vậy x=-5/18

c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}.\frac{4}{3}\)

\(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\).

d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)

\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)

\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)

\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).

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: ...

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

=> \(\left(x+2\right)\cdot5=3\left(2x-1\right)\)

=> \(5x+10=6x-3\)

=> \(6x-5x=10+3\)

=> \(x=13\)

\(\frac{-x}{4}=\frac{-9}{x}\)

=> \(-x^2=4\cdot\left(-9\right)\)

=> \(-x^2=-36\)

=> \(x^2=36\)

=> \(\orbr{\begin{cases}x^2=6^2\\x^2=\left(-6\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

Quỳnh ơi, chuyển 6x sang sẽ là -6x mà viết như cậu phải là -6x+5x :) 

a, \(\frac{x+2}{3}=\frac{2x-1}{5}\)

\(\Leftrightarrow\frac{5x+10}{15}=\frac{6x-3}{15}\Leftrightarrow5x+10=6x-3\Leftrightarrow-x+13=0\Leftrightarrow x=-13\)

b, \(\frac{-x}{4}=\frac{-9}{x}\)\(\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)

\(a,\frac{x+2}{3}=\frac{2x-1}{5}\)

\(< =>\left(x+2\right)5=\left(2x-1\right)3\)

\(< =>5x+10=6x-3\)

\(< =>10+3=6x-5x\)

\(< =>x=13\)( Bạn tú : -x+13=0 <=> x=-13 ?? logic ?! ) 

\(b,\frac{-x}{4}=\frac{-9}{x}\)

\(< =>\frac{x}{4}=\frac{9}{x}\)\(< =>x^2=4.9=36=\pm6^2\)

\(< =>x=\pm6\)

Uả bạn tú nói cái gì vậy !? ko bt chuyển vế đổi dấu à ? 

làm sai thì đừng có đi sủa dạo nhé !! nhìn kĩ đi rồi nói cg không muộn !

Bài làm:

c) \(\left(x-2\right)\left(x+3\right)>0\)

Ta xét 2 trường hợp sau:

+ Nếu \(\hept{\begin{cases}x-2>0\\x+3>0\end{cases}\Rightarrow}\hept{\begin{cases}x>2\\x>-3\end{cases}\Rightarrow}x>2\)

+ Nếu \(\hept{\begin{cases}x-2< 0\\x+3< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x< -3\end{cases}}\Rightarrow x< -3\)

Vậy \(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)

d) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Leftrightarrow3x=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{1}{27}\)

Vậy \(x=\frac{1}{27}\)

Học tốt!!!!