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\(\Leftrightarrow\left(19.75\right):x=\left(\dfrac{33}{5}-\dfrac{51}{16}\right)\cdot\dfrac{35}{6}:\dfrac{5}{2}\)
\(\Leftrightarrow19.75:x=\dfrac{637}{80}\)
hay x=1580/637
a)
ta có \(\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{13}.\dfrac{1}{14}=\dfrac{3}{7}.9.\dfrac{1}{2}.\dfrac{1}{13}-\dfrac{1}{13}.\dfrac{1}{14}\)\(=\dfrac{1}{13}.\left(\dfrac{3}{7}.\dfrac{9}{2}-\dfrac{1}{14}\right)=\dfrac{1}{13}.\dfrac{26}{14}=\dfrac{1.26}{13.14}\)\(=\dfrac{1.13.2}{13.7.2}=\dfrac{1}{7}\)
b)\(x-\left(\dfrac{5}{2}+2x\right)=x-\dfrac{5}{2}-2x=-x-\dfrac{5}{2}=\dfrac{7}{4}\)
\(\Rightarrow-x=\dfrac{7}{4}+\dfrac{5}{2}=\dfrac{17}{4}\)
\(\Rightarrow x=-\dfrac{17}{4}\)(vì -x là số đối của x)
\(\dfrac{x}{2^2}+\dfrac{x}{2^3}+\dfrac{x}{2^4}=\dfrac{x}{3^2}+\dfrac{x}{3^3}+\dfrac{x}{3^4}\)
\(\Leftrightarrow\dfrac{x}{2^2}+\dfrac{x}{2^3}+\dfrac{x}{2^4}-\dfrac{x}{3^2}-\dfrac{x}{3^3}-\dfrac{x}{3^4}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-\dfrac{1}{3^4}\right)=0\)
\(\Leftrightarrow x=0\)
Vậy x = 0
Theo đề ta có:\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{x^2}{64}=\dfrac{y^2}{144}=\dfrac{z^2}{225}\)
Áp dụng t/c của dãy tỉ số = nhau ta có:
\(\dfrac{x^2}{64}=\dfrac{y^2}{144}=\dfrac{z^2}{225}=\dfrac{x^2-y^2}{64-144}=\dfrac{-16}{-80}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{5}\cdot64=\dfrac{64}{5}\\y^2=\dfrac{1}{5}\cdot144=\dfrac{144}{5}\\z^2=\dfrac{1}{5}\cdot225=45\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt{\dfrac{64}{5}};x=-\sqrt{\dfrac{64}{5}}\\y=\sqrt{\dfrac{144}{5}};y=-\sqrt{\dfrac{144}{5}}\\z=\sqrt{45};z=-\sqrt{45}\end{matrix}\right.\)
Vậy............................
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
2.
\(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\) . Ta có : +,ad < bc
\(\Rightarrow\)ad+ab < bc +ab (Cùng thêm ab vào 2 vế)
\(\Rightarrow\)a(b+d) < b(a+c)
\(\Rightarrow\)\(\dfrac{a}{b}\)< \(\dfrac{a+c}{b+d}\)
+, ad < bc
\(\Rightarrow\)ad + cd < bc + cd ( Cùng thêm cd vào 2 vế)
\(\Rightarrow\)d(a+c) < c(b+d)
\(\Rightarrow\)\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) Vậy \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
2.
ta có
\(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\Rightarrow ad< bc\)
xét
\(\dfrac{a}{b}=\dfrac{a\left(b+d\right)}{b\left(b+d\right)}=\dfrac{ab+ad}{b\left(b+d\right)}\)
\(\dfrac{a+c}{b+d}=\dfrac{b\left(a+c\right)}{b\left(b+d\right)}=\dfrac{ab+bc}{b\left(b+d\right)}\)
vì \(\dfrac{ab+ad}{b\left(b+d\right)}< \dfrac{ab+bc}{b\left(b+d\right)}\left(ad< bc\right)\)
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)
xét
\(\dfrac{a+c}{b+d}=\dfrac{d\left(a+c\right)}{d\left(b+d\right)}=\dfrac{ad+cd}{d\left(b+d\right)}\)
\(\dfrac{c}{d}=\dfrac{c\left(b+d\right)}{d\left(b+d\right)}=\dfrac{bc+cd}{d\left(b+d\right)}\)
vì
\(\dfrac{ad+cd}{d\left(b+d\right)}< \dfrac{bc+cd}{d\left(b+d\right)}\left(ad< bc\right)\)
\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\)
từ (1) và (2) => ĐPCM
Ta có :
\(\dfrac{x}{10}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{10}\)
\(\dfrac{y}{2}=\dfrac{z}{3}\Leftrightarrow\dfrac{y}{10}=\dfrac{z}{15}\)
\(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\)
\(\Leftrightarrow\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}=\dfrac{2x-3y+4z}{40-30+60}=\dfrac{330}{70}=\dfrac{33}{7}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{33}{7}\Leftrightarrow x=\dfrac{660}{7}\\\dfrac{y}{10}=\dfrac{33}{7}\Leftrightarrow y=\dfrac{330}{7}\\\dfrac{z}{15}=\dfrac{33}{7}\Leftrightarrow z=\dfrac{495}{7}\end{matrix}\right.\)
Vậy .....
có ngay em ơi:
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{3}\) = \(\dfrac{3}{4}\)
\(\dfrac{2x+1}{3}\) = \(\dfrac{3}{4}\)
2\(x\) + 1 = \(\dfrac{9}{4}\)
2\(x\) = \(\dfrac{9}{4}\) - 1
2\(x\) = \(\dfrac{5}{4}\)
\(x\) = \(\dfrac{5}{4}\) :2
\(x\) = 5/8
\(\dfrac{2}{3}x\) + \(\dfrac{1}{3}\)= \(\dfrac{3}{4}\)
\(\dfrac{2}{3}x=\dfrac{5}{16}\)
\(x=\dfrac{15}{32}\)