\(\frac{1}{\left(x+5\right)\left(x+4\right)}+\frac{1}{\left(x+6\right)\left(x+5\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

 \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{\left(x+7\right)-\left(x+4\right)}{\left(x+7\right)\left(x+4\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\) 

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x=26\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

a: \(\frac{x+1}{2x-6}-\frac{4}{2x-6}\)

\(=\frac{x+1-4}{2\left(x-3\right)}\)

\(=\frac{x-3}{2\left(x-3\right)}=\frac12\)

b: \(\frac{3x-4}{6x+3}-\frac{x-5}{6x+3}\)

\(=\frac{3x-4-x+5}{6x+3}\)

\(=\frac{2x+1}{3\left(2x+1\right)}=\frac13\)

c: \(\frac{x-1}{x-3}-\frac{3x-8}{3-x}+\frac{3-2x}{x-3}\)

\(=\frac{x-1}{x-3}+\frac{3x-8}{x-3}+\frac{3-2x}{x-3}\)

\(=\frac{x-1+3x-8+3-2x}{x-3}=\frac{2x-6}{x-3}=\frac{2\left(x-3\right)}{x-3}\)

=2

d: \(\frac{3}{x+5}-\frac{5}{x-7}\)

\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)

\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)

e: \(\frac{3}{x+5}-\frac{5}{x-7}\)

\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)

\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)

f: \(\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{x^2-4}\)

\(=\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2\left(x+2\right)+3\left(x-2\right)+5x-18}{\left(x-2\right)\left(x+2\right)}=\frac{2x+4+3x-6+5x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{10x-20}{\left(x-2\right)\left(x+2\right)}=\frac{10}{x+2}\)

1.

$(x-2)(x-5)=(x-3)(x-4)$

$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)

Vậy pt vô nghiệm.

2.

$(x-7)(x+7)+x^2-2=2(x^2+5)$

$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$

$\Leftrightarrow -51=10$ (vô lý)

Vậy pt vô nghiệm.

3.

$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$

$\Leftrightarrow 4x+10=-8$

$\Leftrightarrow 4x=-18$

$\Leftrightarrow x=-4,5$

4.

$(x+1)^2=(x+3)(x-2)$

$\Leftrightarrow x^2+2x+1=x^2+x-6$

$\Leftrightarrow x=-7$