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Giải thích các bước giải:

25.ĐKXĐ: $x\ne 2$

$\dfrac1{x-2}+3=\dfrac{3-x}{x-2}$ 

$\to \dfrac{1+3x-6}{x-2}=\dfrac{3-x}{x-2}$

$\to \dfrac{3x-5}{x-2}=\dfrac{3-x}{x-2}$

$\to 3x-5=3-x$

$\to 4x=8$

$\to x=2$ loại

$\to$Phương trình vô nghiệm

26.ĐKXĐ: $x\ne\dfrac32$

$\dfrac{(x+2)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}$

$\to \left(x+2\right)^2-\left(2x-3\right)=x^2+10$

$\to x^2+2x+7=x^2+10$

$\to 2x=3$

$\to x=\dfrac32$

27.ĐKXĐ: $x\ne -3$

$2x-\dfrac{2x^2}{x+3}=\dfrac{4x}{x+3}+\dfrac27$

$\to 14x\left(x+3\right)-14x^2=28x+2\left(x+3\right)$

$\to 42x=30x+6$

$\to 12x=6$

$\to x=\dfrac12$

28.ĐKXĐ: $x\ne -1$

$\dfrac{5x}{2x+2}+1=\dfrac{-6}{x+1}$

$\to 5x+2\left(x+1\right)=-12$

$\to 7x+2=-12$

$\to 7x=-14$

$\to x=-2$

29.ĐKXĐ: $x\ne -1$

$\dfrac2{x+1}+\dfrac1{3x+3}=\dfrac16$

$\to \dfrac2{x+1}+\dfrac1{3(x+1)}=\dfrac16$

$\to \dfrac7{3(x+1)}=\dfrac16$

$\to x+1=14$

$\to x=13$

30.ĐKXĐ: $x\ne 0$

Ta có:

$x+\dfrac1x=x^2+\dfrac1{x^2}$

$\to x^3+x=x^4+1$

$\to x^4-x^3+1-x=0$

$\to x^3(x-1)-(x-1)=0$

$\to (x-1)(x^3-1)=0$

$\to (x-1)^2(x^2+x+1)=0$

Vì $x^2+x+1=(x+\dfrac12)^2+\dfrac34>0$

$\to x-1=0$

$\to x=1$