Giải phương trình:
h) \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
i) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
k) \(\left(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}\right)+\frac{58}{57}+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
l) \(\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-660}{46}=0\)
các bạn không giải thì làm ơn đừng trả lời
\(h.\) \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow\) \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}+4=0\)
\(\Leftrightarrow\) \(\left(\frac{99-x}{101}+1\right)+\left(\frac{97-x}{103}+1\right)+\left(\frac{95-x}{105}+1\right)+\left(\frac{93-x}{107}\right)=0\)
\(\Leftrightarrow\) \(\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow\) \(\left(200-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow\) \(200-x=0\) \(\Leftrightarrow\) \(x=200\)
\(i.\) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\) \(\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{x+116}{4}-4\right)=0\)
\(\Leftrightarrow\) \(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\) \(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\) \(x+100=0\) \(\Leftrightarrow\) \(x=-100\)
\(k.\) \(\left(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}\right)+\frac{58}{57}+2\left(x-1\right)=\frac{2x+7}{3}+\frac{5x-8}{4}\)
\(\Leftrightarrow\) \(\left(1-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}\right)+\frac{58}{57}+2\left(x-1\right)=\frac{4\left(2x+7\right)+3\left(5x-8\right)}{12}\)
\(\Leftrightarrow\) \(\frac{56}{57}+\frac{58}{57}+2\left(x-1\right)=\frac{23x+4}{12}\)
\(\Leftrightarrow\) \(2+2\left(x-1\right)=\frac{23x+4}{12}\)
\(\Leftrightarrow\) \(2x=\frac{23x+4}{12}\)
\(\Leftrightarrow\) \(2x.12=23x+4\)
\(\Leftrightarrow\) \(24x=23x+4\)
\(\Leftrightarrow\) \(x=4\)
Chú ý: \(\frac{8}{1.9}=\frac{9-1}{1.9}=1-\frac{1}{9}\)
Tương tự, ta cũng có: \(\frac{8}{9.17}=\frac{17-9}{9.17}=\frac{1}{9}-\frac{1}{17}\)
\(............\)
\(\frac{8}{49.57}=\frac{57-49}{49.57}=\frac{1}{49}-\frac{1}{57}\)
