b) \(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\) đk x khác 2 và -2

\(\dfrac{\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\)

=>\(x-2-5x-10=2x-3\)

\(-6x=9=>x=\dfrac{3}{2}tm\)

a) \(2\left(x+3\right)=4x-\left(2+x\right)\)

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

\(-x=-8\)=>x=8

\(A=\left(\dfrac{2}{x-2}+\dfrac{4x}{4-x^2}+\dfrac{x}{x+2}\right):\dfrac{1}{x^2+4x+4}\)

\(A=\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{4x}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right).\left(x+2\right)^2\)

\(A=\dfrac{2x+4-4x+x^2-2x}{\left(x-2\right)\left(x+2\right)}.\left(x+2\right)^2\)

\(A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}.\left(x+2\right)^2\)

\(A=\left(x-2\right).\left(x+2\right)=x^2-4\)

xin lỗi mk đọc nhầm đề bài =25 nhé

nếu là 

1 bánh + 1 khoai . 1 cốc =5+2.10=25

17 nhé

\(\dfrac{100}{x}-\dfrac{100}{x+10}=\dfrac{1}{2}\)

=>\(200\left(x+10\right)-200x=x^2+10x\)

=>\(-x^2-10x+2000=0\)

<=>\(-x^2-50x+40x+2000=0\)

<=>\(-x\left(x+50\right)+40\left(x+50\right)=0\)

\(\left(-x+40\right)\left(x+50\right)=0\)

=>\(-x+40=0=>x=40\)

hoặc \(x+50=0=>x=-50\)