Giải thích các bước giải:

Bài 1:

a.$2(3x-1)-3x=10$ 

$\to 6x-2-3x=10$

$\to 3x-2=10$

$\to 3x=12$

$\to x=4$

b.Gửi lại đề

Bài 2:

a.Khi $x=5$

$\to A=\dfrac{5-3}{5+2}=\dfrac27$

b.Ta có:

$B=\dfrac{6-7x}{x^2-4}+\dfrac3{x+2}-\dfrac{x}{2-x}$

$\to B=\dfrac{6-7x}{x^2-4}+\dfrac3{x+2}+\dfrac{x}{x-2}$

$\to B=\dfrac{6-7x}{\left(x+2\right)\left(x-2\right)}+\dfrac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}$

$\to B=\dfrac{6-7x+3\left(x-2\right)+x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}$

$\to B=\dfrac{x^2-2x}{\left(x+2\right)\left(x-2\right)}$

$\to B=\dfrac{x}{x+2}$

c.Ta có:

$AB=1$

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

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

$\to x^2-3x=x^2+4x+4$

$\to 7x=-4$

$\to x=-\dfrac47$