Question

[Sarah Garritsen]

Cho biểu thức M = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

a) Rút gọn M.

b) Tìm giá trị của x để M = \(\frac{1}{3}\)

Answer 1

[l҉o҉n҉g҉ d҉z҉]

\(M=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

a) ĐKXĐ : x ≠ -3 , x ≠ 2

\(=\frac{x+2}{x+3}-\frac{5}{x^2-2x+3x-6}-\frac{1}{x-2}\)

\(=\frac{x+2}{x+3}-\frac{5}{x\left(x-2\right)+3\left(x-2\right)}-\frac{1}{x-2}\)

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

\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)

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

b) Để M = 1/3

=> \(\frac{x-4}{x-2}=\frac{1}{3}\)( x ≠ -3 , x ≠ 2 )

=> 3( x - 4 ) = x - 2

=> 3x - 12 - x + 2 = 0

=> 2x - 10 = 0

=> 2x = 10

=> x = 5 ( tm )

Vậy x = 5 thì M = 1/3

Answer 2

[Đặng Ngọc Quỳnh]

đk: \(x\ne2,x\ne-3\)

a) Ta có: \(M=\frac{-4+x^2}{x^2+x-6}-\frac{5}{x^2+x-6}-\frac{x+3}{x^2+x-6}\)

\(=\frac{x^2-x-12}{x^2+x-6}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)

b) \(M=\frac{1}{3}\Rightarrow\frac{x-4}{x-2}=\frac{1}{3}\Leftrightarrow3x-12=x-2\Leftrightarrow x=5\)