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Xét : 1/8 - 1/52 + 1/68
= 3/4 . 1/6 - 3/4 . 1/39 + 3/4 . 1/68
= 3/4 . (1/6-1/39+1/51)
=> E = 1/(3/4) = 4/3
Tk mk nha
\(\text{E}=\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
\(\text{E}=\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{3}{4}.\frac{1}{6}-\frac{3}{4}.\frac{1}{39}+\frac{3}{4}.\frac{1}{68}}\)
\(\text{E}=\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{3}{4}\left(\frac{1}{6}-\frac{1}{39}+\frac{1}{51}\right)}\)
\(\text{E}=\frac{1}{\left(\frac{3}{4}\right)}=\frac{4}{3}\)
a) 31/23 - ( 7/32 + 8/22)
= 31/23 - 7/32 + 8/23
= ( 31/23 + 8/23 ) - 7/32
= 32/22 - 7/32
= 39/32
Ccá ý khác làm tương tự
Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
a) \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}=1-\frac{7}{32}=\frac{25}{32}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}-\left(\frac{79}{67}-\frac{12}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1=\frac{1}{3}\)
d) \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}=\frac{1}{7}.\left(\frac{1}{3}-\frac{1}{3}\right)+\frac{17}{19}=\frac{17}{19}\)
e) \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}=\frac{7}{5}.\left(\frac{3}{9}+\frac{2}{9}\right)=\frac{7}{5}.\frac{5}{9}=\frac{7}{9}\)
