a) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\)\(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(\Leftrightarrow20x^2-20x^2+4x+30x-25x-8x=10-6\)

\(\Leftrightarrow x=4\)

b) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow15x^2-15x^2-102x-5x+120x+125x=1000-34\)

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=7\)

a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(4x+30x-25x-8x=10-6\)

\(x=4\)

b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow-102x-5x-120x-125x=1000-34\)

\(\Leftrightarrow-352x=966\)

\(\Leftrightarrow x=-\frac{483}{176}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2.\left(2x+3\right)-\left(4x+5\right)}{2.\left(5x+2\right)-\left(10x+2\right)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)

Suy ra:

\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow2.\left(2x+3\right)=1.\left(5x+2\right)\Rightarrow4x+6=5x+2\)

\(\Rightarrow x=4\)

\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)

\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)

\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)

\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)

chỗ cuối tớ sai 

\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)

đây nha , e xin lỗi

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

                                                          \(=\frac{3\left(x-1\right)+\left(2x-1\right)-2.2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)

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

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

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

b) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

                                                   \(=\frac{3x.10\left(x-y\right)-x.5\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{30x\left(x-y\right)+5x\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left[6\left(x-y\right)-\left(x+y\right)\right]}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left(5x-7y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{x\left(5x-7y\right)}{10\left(x-y\right)\left(x+y\right)}\)

c) \(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}=\frac{5x^2-y-x\left(3x-2y\right)}{xy}\)

                                                \(=\frac{5x^2-y-3x^2+2xy}{xy}\)

                                               \(=\frac{2x^2-y+2xy}{xy}\)

d) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)

                                            \(=\frac{3x-x+6}{2x\left(x+3\right)}\)

                                            \(=\frac{2x+6}{2x\left(x+3\right)}\)

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

                                            \(=\frac{2}{2x}=\frac{1}{x}\) 

a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)

=> 2x + 7 = 4 

     2x        = 4 - 7 

     2x        = -3

       x        = -3 : 2

       x         = -1,5

   Vậy x = -1,5

a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(\Leftrightarrow4x+30x-25x-8x=10-6\)

\(\Leftrightarrow1x=4\)

\(\Leftrightarrow x=4:1\)

\(\Leftrightarrow x=4\)

b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=966:138\)

\(\Leftrightarrow x=7\)

a. \(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(20x^2+34x+5=20x^2+33x+10\)

rút gọn \(x=5\)

mình gửi câu a tí nữa mình gửi câu b

a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)

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

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

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

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

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

\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)

tương tự đến hết nha a hay cj gì đps ! 

a) \(\frac{4.x+1}{2}-\frac{3.x+2}{3}=\frac{3.\left(4.x+1\right)-2.\left(3.x+2\right)}{6}\)

                                                \(=\frac{12.x+3-6.x-4}{6}\)

                                                   \(=\frac{6.x-1}{6}\)

b)\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)

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

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

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

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

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

\(=\frac{x+1}{x.\left(x-1\right)}\)

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

c)\(\frac{3}{2.x^2+2.x}+\frac{2.x-1}{x^2-1}-\frac{1}{2}\)

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

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

\(=\frac{3.x-3+4.x^2-2.x-x.\left(x^2-1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{3.x-3+4.x^2-2.x-x^3+x}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{2.x-3+4.x^2-x^3}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-x^3+4.x^2+2.x-3}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-x^3-x^2+5.x^2+5.x-3.x-3}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-x^2.\left(x+1\right)+5.x.\left(x+1\right)-3.\left(x+1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-\left(x+1\right).\left(x^2-5.x+3\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-\left(x^2-5.x+3\right)}{2.x.\left(x-1\right)}\)

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

Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)

\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)

\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)

Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)

\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)

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

\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)

a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)

\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)

\(=\frac{3x^2-3y^2}{50}\)

c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)

\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)

\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)

\(=\frac{x+y+2}{y-x}\)

d) \(\frac{\frac{x-y}{x+y}-\frac{x+y}{x-y}}{1-\frac{x^2}{x^2+y^2}}\)

\(=\frac{\frac{x^2-2xy+y^2}{x^2-y^2}-\frac{x^2+2xy+y^2}{x^2-y^2}}{\frac{y^2}{x^2+y^2}}\)

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

\(=\frac{2x^2+2y^2}{x^2-y^2}.\frac{x^2+y^2}{y^2}\)

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