cứu mị :<

\(=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}\cdot\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)

\(\dfrac{x^2+x}{5x^2+10x+5}:\dfrac{3x+3}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x^2+2x+1\right)}:\dfrac{3\left(x+1\right)}{5x-1}=\dfrac{x\left(x+1\right)}{5\left(x+1\right)^2}.\dfrac{5x-1}{3\left(x+1\right)}=\dfrac{x\left(5x-1\right)}{15\left(x+1\right)^2}\)

\(\dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}=\dfrac{6}{10\left(x-1\right)}-\dfrac{x+5}{10\left(x-1\right)}=\dfrac{6-\left(x+5\right)}{10\left(x-1\right)}\)

\(=\dfrac{-\left(x-1\right)}{10\left(x-1\right)}=-\dfrac{1}{10}\)

   \(\dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}\)

= \(\dfrac{3}{5.(x-1)}\) - \(\dfrac{x+5}{2.5.(x-1)}\) 

= \(\dfrac{6-x-5}{10.(x-1)}\)

= \(\dfrac{1-x}{10.(x-1)}\)

= \(\dfrac{-1(x-1)}{10(x-1)}\)

= \(\dfrac{-1}{10}\)

\(đk:x\ne1\\ \dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}\\ =\dfrac{3.2}{2.\left(5x-5\right)}-\dfrac{x+5}{10x-10}\\ =\dfrac{6}{10x-10}-\dfrac{x+5}{10x-10}\\ =\dfrac{6-x-5}{10x-10}\\ =\dfrac{1-x}{10x-10}\\ =\dfrac{-\left(x-1\right)}{10\left(x-1\right)}\\ =\dfrac{-1}{10}\)

a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)

mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)

nên \(x^2-10x-2000=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)

Vậy: S={-40;50}

\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

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

dkxd : x ≠ 0

          x ≠ 5

          x ≠ -5

MTC : 2x(x - 5)(x + 5)

Quy đồng mẫu thức hai vế của phương trình :

⇒ \(\dfrac{2\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}\) = \(\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)

Suy ra : 2(x - 5)(x + 5) - (x - 5)(x + 5) = x(x + 25)

         \(\Leftrightarrow\) 2(x² - 25) - (x² - 25) = x² + 25x

         \(\Leftrightarrow\) 2x² - 50 - x² + 25 - x² - 25x = 0

        \(\Leftrightarrow\) -25 - 25x = 0

        \(\Leftrightarrow\) -25x = 25

        \(\Leftrightarrow\) x = \(\dfrac{25}{-25}=-1\) (thỏa mãn)

 Vậy S = \(\left\{-1\right\}\)

 Chúc bạn học tốt

Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

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

Suy ra: \(2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)

\(\Leftrightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

hay \(x=-\dfrac{5}{3}\)(thỏa ĐK)

\(x\ne0;x\ne\pm5\)

PT \(\Leftrightarrow\dfrac{x+25}{2\left(x+5\right)\left(x-5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}=0\)

\(\Rightarrow x^2+25x-2x^2-20x-50+x^2-10x+25=0\)

\(\Leftrightarrow-5x-25=0\)

\(\Leftrightarrow x=-5\) (ktm)
Vậy pt vô nghiệm.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\).

\(PT\Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\)

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

\(\Rightarrow x\left(x+25\right)-2\left(x+5\right)^2=\left(5-x\right)\left(x-5\right)\)

\(\Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=10x-x^2-25\)

\(\Leftrightarrow-5x=25\Leftrightarrow x=-5\) (loại)

Vậy PT vô nghiệm

\(\dfrac{x+1}{1}+\dfrac{2x+3}{3}+\dfrac{3x+5}{5}+...+\dfrac{10x+19}{19}=12+\dfrac{4}{3}+\dfrac{6}{5}+...+\dfrac{20}{19}\)

\(x+1+\dfrac{2x}{3}+1+\dfrac{3x}{5}+1+...+\dfrac{10x}{19}+1-12-\dfrac{4}{3}-\dfrac{6}{5}-...-\dfrac{20}{19}=0\)

\(x+\dfrac{2x}{3}-\dfrac{4}{3}+\dfrac{3x}{5}-\dfrac{6}{5}+...+\dfrac{10x}{19}-\dfrac{20}{19}+10-12=0\)

\(x-2+\dfrac{2x-4}{3}+\dfrac{3x-6}{5}+...+\dfrac{10x-20}{19}=0\)

\(x-2+\dfrac{2\left(x-2\right)}{3}+\dfrac{3\left(x-2\right)}{5}+...+\dfrac{10\left(x-2\right)}{19}=0\)

\(\left(x-2\right)\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)=0\)

Ta thấy \(\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)>0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)