B1:
a) A = \(\dfrac{1}{x+2}+\dfrac{x^2-x-2}{x^2-7x+10}-\dfrac{2x-4}{x-5}\)
= \(\dfrac{1}{x+2}+\dfrac{\left(x^2-2x\right)+\left(x-2\right)}{\left(x^2-2x\right)-\left(5x-10\right)}-\dfrac{2\left(x-2\right)}{x-5}\)
= \(\dfrac{1}{x+2}+\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{2\left(x-2\right)}{x-5}\) [ĐKXĐ: x ≠ -2; x ≠ 5]
= \(\dfrac{x-5}{\left(x+2\right)\left(x-5\right)}+\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x-5\right)\left(x+2\right)}\)
= \(\dfrac{-x^2+4x+5}{\left(x+2\right)\left(x-5\right)}\)
= \(\dfrac{-x\left(x-5\right)-\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}\)
= \(\dfrac{\left(x-5\right)\left(-x-1\right)}{\left(x-5\right)\left(x+2\right)}\)
= \(-\dfrac{x+1}{x+2}\)
b) Thay x = 3 vào A, ta có:
A = \(-\dfrac{3+1}{3+2}=-\dfrac{4}{5}\)
c) A = 1
<=> \(-\dfrac{x+1}{x+2}\)= 1 <=> -(x + 1) = x + 2 <=> -x - 1 = x + 2
<=> -2x = 3 <=> x = \(\dfrac{-3}{2}\)
d) A = \(\dfrac{-\left(x+1\right)}{x+2}\)= \(\dfrac{-\left(x+2\right)+1}{x+2}\)= -1 + \(\dfrac{1}{x+2}\)
A đạt giá trị nguyên khi 1 chia hết cho x + 2 hay x + 2 ∈ Ư(1) = {1;-1}
* x + 2 = 1 <=> x = -1
* x + 2 = -1 <=> x = -3
B2: M = \(\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
= \(\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{5\left(10-x\right)}{2x\left(x+5\right)}\)[ĐKXĐ: x ≠ 0; x ≠ -5
= \(\dfrac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)
= \(\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
= \(\dfrac{x^2+4x-5}{2\left(x+5\right)}\)
= \(\dfrac{\left(x^2+5x\right)-\left(x+5\right)}{2\left(x+5\right)}\)
\(\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}=\dfrac{x-1}{2}\)
b) Thay x = 3 vào M, ta có:
M = \(\dfrac{3-1}{2}=1\)
Thay x = 5 vào M, ta có:
M = \(\dfrac{5-1}{2}=2\)
