a) \(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)
\(\Rightarrow\left(1+\frac{59-x}{41}\right)+\left(1+\frac{57-x}{43}\right)+\left(1+\frac{55-x}{45}\right)+\left(1+\frac{53-x}{47}\right)+\left(1+\frac{51-x}{49}\right)=0\)
\(\Rightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
Mà \(\frac{1}{41}+\frac{1}{42}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
Vậy x = 100
a) x=100
b) x ≈ 99.7784191416069
c)x=-100