A ) / X + 2 / + / X + / + / X + \(\frac{1}{2}\)/ = 4X
b ) / x + 1,1 / + / x + 1,2 / + / x + 1,3 / + / x + 1,4 / = 5x
c ) / x + \(\frac{1}{1.3}\)/ + / x + \(\frac{1}{3.5}\)/ + / x +\(\frac{1}{5.7}\) / + ........ + / x + \(\frac{1}{97.99}\)/ = 50x
d ) / x + \(\frac{1}{1.5}\)/ + / x + \(\frac{1}{5.9}\)/ + / x + \(\frac{1}{9.13}\)/ + ......... + / x + \(\frac{1}{397.401}\)/ = 101x
Bài này khá ez thôi:
a) bạn sửa lại đề rồi làm theo cách làm của b,c,d nhé
b) Ta có: \(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\left(\forall x\right)\)
\(\Rightarrow5x\ge0\Rightarrow x\ge0\) khi đó:
\(PT\Leftrightarrow x+1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow x=5\)
c,d tương tự nhé
c,\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}+\right|+...+\left|x+\frac{1}{97.99}\right|\ge0\forall x\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)Khi đó:
\(x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)
\(\Rightarrow49x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=50x\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{49}{99}\)
