GIÚP MÌNH VỚI
2/1×4 + 2/4×7 + 2/7×9+....+ 2/196×199
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a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
\(d,\dfrac{1}{2}+\dfrac{4}{3}=\dfrac{3}{6}+\dfrac{8}{6}=\dfrac{3+8}{6}=\dfrac{11}{6}\)
\(e,\dfrac{6}{7}+\dfrac{11}{4}=\dfrac{24}{28}+\dfrac{77}{28}=\dfrac{24+77}{28}=\dfrac{101}{28}\)
\(g,\dfrac{9}{5}+\dfrac{7}{4}=\dfrac{36}{20}+\dfrac{35}{20}=\dfrac{36+35}{20}=\dfrac{71}{20}\)
\(h,\dfrac{2}{8}+\dfrac{1}{4}=\dfrac{2}{8}+\dfrac{2}{8}=\dfrac{2+2}{8}=\dfrac{4}{8}=\dfrac{1}{2}\)
\(i,\dfrac{3}{20}+\dfrac{2}{5}=\dfrac{3}{20}+\dfrac{8}{20}=\dfrac{3+8}{20}=\dfrac{11}{20}\)
\(k,\dfrac{6}{35}+\dfrac{2}{7}=\dfrac{6}{35}+\dfrac{10}{35}=\dfrac{6+10}{35}=\dfrac{16}{35}\)
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)
\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)
\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))
\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)
\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.9}+...+\frac{2}{196.199}=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.9}+...+\frac{3}{196.199}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{196}-\frac{1}{199}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{199}\right)\)
\(=\frac{2}{3}.\frac{198}{199}\)
\(=\frac{132}{199}\)