Tìm các số nguyên dương x biết:
a)\(\dfrac{19}{-23}\)\(< \dfrac{-19}{x}\)\(< \dfrac{19}{-29}\) b)\(\dfrac{2}{3}\)\(< \dfrac{88}{x}\)\(< \dfrac{11}{16}\) c)\(\dfrac{4}{x}\)\(< \dfrac{x}{8}\)\(< \dfrac{5}{x}\)
A=\(-\dfrac{68}{123}\)x\(-\dfrac{23}{79}\)
B=\(-\dfrac{14}{79}\)x\(-\dfrac{68}{7}\)x\(-\dfrac{46}{123}\)
C=\(-\dfrac{4}{19}\)x\(-\dfrac{3}{19}\)x\(-\dfrac{2}{19}\) ... \(\dfrac{2}{19}\)x\(\dfrac{3}{19}\)x\(\dfrac{4}{19}\)
a)So sánh A,B,C
b)Tính B:A
a) Ta có:
\(A=\dfrac{-68}{123}\cdot\dfrac{-23}{79}=\dfrac{68}{123}\cdot\dfrac{23}{79}\)
\(B=\dfrac{-14}{79}\cdot\dfrac{-68}{7}\cdot\dfrac{-46}{123}=-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)\)
\(C=\dfrac{-4}{19}\cdot\dfrac{-3}{19}\cdot...\cdot\dfrac{0}{19}\cdot...\cdot\dfrac{3}{19}\cdot\dfrac{4}{19}=0\)
Suy ra A là số hữu tỉ dương, B là số hữu tỉ âm và C là 0.
Vậy A > C > B.
b) Ta có:
\(\dfrac{B}{A}=\dfrac{-\left(\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\right)}{\dfrac{68}{123}\cdot\dfrac{23}{79}}=-\dfrac{14}{79}\cdot\dfrac{68}{7}\cdot\dfrac{46}{123}\cdot\dfrac{123}{68}\cdot\dfrac{79}{23}\)
\(\dfrac{B}{A}=-\dfrac{14\cdot68\cdot46\cdot123\cdot79}{79\cdot7\cdot123\cdot68\cdot23}=-\left(2\cdot2\right)=-4\)
Vậy B : A = -4
bài 45: Tìm x, biết:
a) x=\(\dfrac{-1}{2}+\dfrac{3}{4}\); b) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
a, x= \(\dfrac{1}{4}\)
b, x= \(\dfrac{1}{5}\)
x=-2/4+3/4 x=1/4
6x/30=25/30+-19/30 6x/30=6/30 6x=6 x=1
Giải phương trình:\(\dfrac{x-29}{1970}+\dfrac{x-27}{1972}+\dfrac{x-25}{1974}+\dfrac{x-23}{1976}+\dfrac{x-21}{1978}+\dfrac{x-19}{1980}=\dfrac{x-1970}{29}+\dfrac{x-1972}{27}+\dfrac{x-1974}{25}+\dfrac{x-1976}{23}+\dfrac{x-1978}{21}+\dfrac{x-1980}{19}\)
Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}\)\(=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
\(\Leftrightarrow\left(\frac{x-29}{1970}-1\right)+\left(\frac{x-27}{1972}-1\right)+\left(\frac{x-25}{1974}-1\right)+\left(\frac{x-23}{1976}-1\right)+\left(\frac{x-21}{1978}-1\right)+\left(\frac{x-19}{1980}-1\right)\)\(=\left(\frac{x-1970}{29}-1\right)+\left(\frac{x-1972}{27}-1\right)+\left(\frac{x-1974}{25}-1\right)+\left(\frac{x-1976}{23}-1\right)+\left(\frac{x-1978}{21}-1\right)+\left(\frac{x-1980}{19}-1\right)\)
\(\Leftrightarrow\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1999}{1978}+\frac{x-1999}{1980}\)\(=\frac{x-1999}{29}+\frac{x-1999}{27}+\frac{x-1999}{25}+\frac{x-1999}{24}+\frac{x-1999}{21}+\frac{x-1999}{19}\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}\right)\)\(=\left(x-1999\right)\left(\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}-\frac{1}{23}-\frac{1}{21}-\frac{1}{19}\right)=0\)\(\Leftrightarrow\) \(x-1999=0\) (Vì ...khác 0)
\(\Leftrightarrow x=1999\)(thỏa mãn)
Vậy \(x=1999\)
\(A=\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{65}+...+\dfrac{1}{99}=\dfrac{16}{x}\)
\(B=\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{x}{16}\)\(C=\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{9}\right)\times\left(1-\dfrac{1}{16}\right)\times...\times\left(1-\dfrac{1}{100}\right)=\dfrac{22}{x}\)
m.ng giúp em với
a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)
=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11
=>32/x=1/3-1/11=8/33
=>x=32:8/33=132
b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)
=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16
=>x=90
c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)
=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10
=>22/x=1/10*11/2=11/20=22/40
=>x=40
\(\dfrac{19}{8}\)x\(\dfrac{16}{9}\)+\(\dfrac{19}{8}\)x\(\dfrac{2}{9}\)-\(\dfrac{19}{8}\)
\(=\dfrac{19}{8}.\dfrac{16}{9}+\dfrac{19}{8}.\dfrac{2}{9}-\dfrac{19}{8}.1=\dfrac{19}{8}.\left(\dfrac{16}{9}+\dfrac{2}{9}-1\right)\)
\(=\dfrac{19}{8}.1=\dfrac{19}{8}\)
\(\dfrac{19}{8}\times\dfrac{16}{9}+\dfrac{19}{8}\times\dfrac{2}{9}-\dfrac{19}{8}\)
\(=\dfrac{19}{8}\times\left(\dfrac{16}{9}+\dfrac{2}{9}-1\right)\)
\(=\dfrac{19}{8}\times1\)
\(=\dfrac{19}{8}\)
Tìm số nguyên x thỏa mãn điều kiện
\(\dfrac{-19}{23}\) < \(\dfrac{7}{x}\) < \(\dfrac{-19}{25}\)
a). \(C=\dfrac{x^4+x^8+x^{12}+x^{16}+x^{20}+x^{24}+x^{28}+1}{x^3+x^7+x^{11}+x^{15}+x^{19}+x^{23}+x^{27}+x^{31}}\)
b). \(F=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{2011.2012.2013.2014}\)
c). \(\dfrac{14044}{12345}=1+\dfrac{1}{7+\dfrac{1}{8+\dfrac{1}{9+\dfrac{1}{x+\dfrac{1}{y}}}}}\)
\(a.C=\dfrac{x^4+x^8+x^{12}+x^{16}+x^{20}+x^{24}+x^{28}+1}{x^3+x^7+x^{11}+x^{15}+x^{19}+x^{23}+x^{27}+x^{31}}=\dfrac{x^{28}+x^{24}+...+x^8+x^4+1}{x^3\left(x^{28}+x^{24}+...+x^8+x^4+1\right)}=\dfrac{1}{x^3}\) Tại x = 2015 thì : \(C=\dfrac{1}{x^3}=\dfrac{1}{2015^3}\)
\(b.F=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{2011.2012.2013.2014}\)
\(3F=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{2011.2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{3.4.5}-\dfrac{1}{4.5.6}+...+\dfrac{1}{2011.2012.2013}-\dfrac{1}{2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2012.2013.2014}\)
Tới đây dễ rồi , bạn tự tính nốt .
*Thực hiện
1/ (\(\dfrac{2021}{2020}\)+\(\dfrac{2020}{2021}\)) x (\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\))
2/ (\(\dfrac{7}{19}\)-\(\dfrac{5}{12}\)):\(\dfrac{-5}{8}\)-(\(\dfrac{7}{19}\)-\(\dfrac{29}{12}\)):\(\dfrac{5}{8}\)
3/ \(\dfrac{-5}{6}\)x\(\dfrac{7}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{14}{24}\)-\(\dfrac{5}{6}\)x\(\dfrac{3}{24}\)
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
mink chịu bài này nó rất khó
1.Thực hiện các phép tính sau :
a) \(-\dfrac{4}{3}.\dfrac{5}{12}+\dfrac{1}{3}.\dfrac{5}{12}\) b)\(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)
2.Tìm x, biết:
a) 2x+19=\(^{5^2}\) b)\(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)
1,
a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)
b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)
=\(\dfrac{-2}{15}\)
2,
a, 2x+19=25
=>x=3
b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)
=>x=\(\dfrac{-3}{2}\)
Bài 1:
a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)
\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)
\(=\dfrac{-5}{12}\)
b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)
\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)
1)
a) \(-\dfrac{4}{3}.\dfrac{5}{12}+\dfrac{1}{3}.\dfrac{5}{12}=\dfrac{5}{12}.\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)=\dfrac{5}{12}.\left(-1\right)=-\dfrac{5}{12}\)
b) \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right).\dfrac{28}{3}=3+\dfrac{1}{5}-\dfrac{45}{14}.\dfrac{28}{3}\)
\(=3+\dfrac{1}{5}-30=-27+\dfrac{1}{5}=-\dfrac{134}{5}\)
2)
a) \(2x+19=25\)
\(2x=25-19=6\)
\(x=3\)
b) \(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)
\(-\dfrac{2x}{9}=\dfrac{4}{21}+\dfrac{1}{7}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}.\left(-\dfrac{9}{2}\right)=-\dfrac{3}{2}\)