Question

a) Tìm x thỏa mãn: \(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{5}=0\)

b) Tìm số nguyên x,y biết: 42 - 3|y - 3| = 4(2012 - x)⁴

Answer 1

\(\dfrac{x-10}{30}+\dfrac{x-14}{43}+\dfrac{x-5}{95}+\dfrac{x-148}{8}=0\\ \Rightarrow\left(\dfrac{x-10}{30}-3\right)+\left(\dfrac{x-14}{43}-2\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-148}{8}+6\right)=0\\ \Rightarrow\dfrac{x-100}{30}+\dfrac{x-100}{43}+\dfrac{x-100}{95}+\dfrac{x-100}{8}=0\\ \Rightarrow x-100=0\\ \Rightarrow x=100\)

Answer 2

\(4\left(2012-x\right)^4\ge0\\ \Rightarrow42-3\left|y-3\right|\ge0\\ \Rightarrow3\left|y-3\right|\le42\\ \Rightarrow\left|y-3\right|\le14\\ \Rightarrow-14\le y-3\le14\\ \Rightarrow-11\le y\le17\)

Vậy x=2012 và -11<=y<=17