a: ĐKXĐ: x<>-3/2
\(\frac{x^2+3x+2}{2x+3}=\frac{2x-5}{4}\)
=>\(\left(2x+3\right)\left(2x-5\right)=4\left(x_{}^2+3x+2\right)\)
=>\(4x^2-10x+6x-15=4x^2+12x+8\)
=>-4x-15=12x+8
=>-16x=23
=>\(x=-\frac{23}{16}\) (nhận)
b: ĐKXĐ: x∉{3;-3}
Ta có: \(\frac{2x+3}{x-3}-\frac{4}{x+3}=\frac{24}{x^2-9}+2\)
=>\(\frac{2x+3}{x-3}-\frac{4}{x+3}=\frac{24}{\left(x-3\right)\left(x+3\right)}+2\)
=>\(\frac{\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{24+2\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}\)
=>\(\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)-24-2\left(x^2-9\right)=0\)
=>\(2x^2+6x+3x+9-4x+12-24-2x^2+18=0\)
=>5x+15=0
=>5x=-15
=>x=-3(loại)
c: ĐKXĐ:x>=5/3
\(\sqrt{3x-5}=3\)
=>\(3x-5=3^2=9\)
=>3x=14
=>x=14/3(nhận)
d: ĐKXĐ:x>=-5/2
\(\sqrt{2x+5}=2\)
=>\(2x+5=2^2=4\)
=>2x=4-5=-1
=>\(x=-\frac12\) (nhận)
