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:

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

\(\Rightarrow x=4\)

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

=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x²+4x+30x+6=20x²+25x+8x+10

=>20x²-20x²+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

B)=>(3x-1).(5x-34)=(40-5x).(25-3x)

=>15x²-102x-5x+34=1000-120x-125x+15x²

=>15x²-107x+34=1000-245x+15x²

=>15x²-15x²-107x+245x=1000-34

=>0-107x+245x=966

=>138x=966

=>x=7

A,=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x2+4x+30x+6=20x2+25x+8x+10

=>20x2-20x2+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

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/ 12-3(x-2)=(x+2)(1-3x)+2x

\(\Leftrightarrow18-3x=-3x^2-3x+2\)

\(\Leftrightarrow3x^2=-16\left(vl\right)\)

=> phương trình vô nghiệm

b/\(\left(x+5\right)\left(x+2\right)\) =3(4x-2)+(x-5)

\(\Leftrightarrow x^2+3x+10=13x-11\)

\(\Leftrightarrow x^2-10x+21=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

c/\(\frac{x-5}{x^2-5x}-\frac{x-5}{2x^2-10x}=\frac{x+25}{2x^2-50}\)(x khác 0)

\(\Leftrightarrow\frac{x-5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x-5\right)}=\frac{x^2+25}{2x^2-50}\)

\(\frac{\Leftrightarrow1}{x}-\frac{1}{2x}=\frac{x+25}{2x^2-50}\)

\(\Leftrightarrow\frac{1}{2x}=\frac{x+25}{2x^2-50}\Leftrightarrow2x^2-50=2x^2+50x\)

\(\Leftrightarrow50x=-50\Leftrightarrow x=-1\)(tm)

d/4x²-1=(2x+1)(3x-5)

\(\Leftrightarrow4x^2-1=6x^2-7x-5\)

\(\Leftrightarrow2x^2-7x-4=0\Leftrightarrow\left(x-4\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\frac{1}{2}\end{matrix}\right.\)

e/ \(x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)