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a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
Ta có :
\(\dfrac{a}{b}=\dfrac{a.\left(b+d\right)}{b.\left(b+d\right)}=\dfrac{ab+bd}{b^2+bd}\)
\(\dfrac{a+c}{b+d}=\dfrac{b\left(a+c\right)}{b\left(b+d\right)}=\dfrac{ab+bc}{b^2+bd}\)
Ta so sánh :
\(\dfrac{ab+bd}{b^2+bd}\) và \(\dfrac{ab+bc}{b^2+bd}\)
Vì cùng mẫu nên ta chỉ so sánh :
\(ab+bd\) và \(ab+bc\)
\(\Rightarrow\) Ta tiếp tục so sánh :
\(bd\) và bc thì ta có : bd < bc (1)
Từ 1, suy ra :
\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
Mà \(\dfrac{a}{b}< \dfrac{c}{d}\)
Suy ra : \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\) (đpcm)
tính chất trên gọi là tính chất bắc cầu, ta so sánh hai phân số với một số (phân số) thứ 3.
Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
a: \(\frac{23}{55}=1-\frac{32}{55};\frac{1978}{2010}=1-\frac{32}{2010}\)
Ta có: 55<2010
=>\(\frac{32}{55}>\frac{32}{2010}\)
=>\(-\frac{32}{55}<-\frac{32}{2010}\)
=>\(-\frac{32}{55}+1<-\frac{32}{2010}+1\)
=>\(\frac{23}{55}<\frac{1978}{2010}\)
b: \(\frac{2003\cdot2004-1}{2003\cdot2004}=1-\frac{1}{2003\cdot2004}\)
\(\frac{2004\cdot2005-1}{2004\cdot2005}=1-\frac{1}{2004\cdot2005}\)
Ta có: \(2003\cdot2004<2004\cdot2005\)
=>\(\frac{1}{2003\cdot2004}>\frac{1}{2004\cdot2005}\)
=>\(\frac{-1}{2003\cdot2004}<\frac{-1}{2004\cdot2005}\)
=>\(\frac{-1}{2003\cdot2004}+1<\frac{-1}{2004\cdot2005}+1\)
=>\(\frac{2003\cdot2004-1}{2003\cdot2004}<\frac{2004\cdot2005-1}{2004\cdot2005}\)
c: \(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)
\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\)
=>\(\frac{a}{b}>\frac{a+m}{b+m}\)
d: a<b
=>a-b<0
\(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+am-ab-bm}{b\left(b+m\right)}\)
\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}<0\)
=>\(\frac{a}{b}<\frac{a+m}{b+m}\)