\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\left(DK:x\ne-\frac{2}{5};x\ne-\frac{1}{5}\right)\)
\(\Rightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\Rightarrow20x^2+34x+6=20x^2+33x+10\Rightarrow x=4\)(thoả mãn)
Vậy x = 4
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)
\(\Leftrightarrow2x\left(10x+2\right)+3\left(10x+2\right)=5x\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+20x+6=20x^2+25x+9x+10\)
\(\Leftrightarrow20x^2+4x+20x+6-\left(20x^2+25x+9x+10\right)=0\)\(\Rightarrow20x^2+24x+6-\left(20x^2+34x+10\right)=0\)
\(\Leftrightarrow-10x-4=0\)
\(\Leftrightarrow-10x=4\)
\(\Leftrightarrow x=-\frac{4}{10}\)
Mình nhầm
2x+3/5x+2=4x+5/10x+2
=> (2x+3)(10x+2)=(5x+2)(4x+5)
=> 20x^2+4x+30x+6=10x^2+25x+8x+10 ( Vì cả hai vế đều có 10x^2 nên ta xóa đi )
=> 34x+6=33x+10
=> 34x-33x=-6+10
=> x=4
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\Leftrightarrow\frac{\left(2x+3\right)\left(10x+2\right)}{\left(5x+2\right)\left(10x+2\right)}=\frac{\left(4x+5\right)\left(5x+2\right)}{\left(10x+2\right)\left(5x+2\right)}\)\(\Leftrightarrow\frac{20x^2+34x+6}{50x^2+30x+4}=\frac{20x^2+33x+10}{50x^2+30x+4}\)
\(\Leftrightarrow20x^2+34x+6=20x^2+33x+10\Leftrightarrow34x+6=33x+10\)
\(\Leftrightarrow34x+6-\left(33x+10\right)=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)
=>(2x+3).(4x+5)=(4x+5).(5x+2)
=>20x^+30x+4x+6=20x^2+25x+8x+10
=>20x^2+34x+6=20x^2+33x+10
=>1x=4 ( 2 vế bớt 20x^2 và 6 đơn vị)
Vậy x=4
