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a) \(\dfrac{-12}{15}+\dfrac{-4}{26}=\dfrac{-4}{5}+\dfrac{-2}{13}=\dfrac{-52-10}{65}=\dfrac{-62}{65}\)
b) \(5\dfrac{1}{3}-2\dfrac{4}{5}=\dfrac{16}{3}-\dfrac{14}{5}=\dfrac{80}{15}-\dfrac{42}{15}=\dfrac{38}{15}\)
c) \(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)+\dfrac{-5}{10}=\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{1}{2}=\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{35}{70}=\dfrac{41}{70}\)
d) \(-1\dfrac{2}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-9}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-54}{42}+\dfrac{9}{42}-\dfrac{10}{42}=\dfrac{-55}{42}\)
e) \(12-\dfrac{11}{121}+\left(\dfrac{-8}{9}\right)-\left(-\dfrac{3}{7}\right)\)
\(=12-\dfrac{11}{121}-\dfrac{8}{9}+\dfrac{3}{7}\)
\(=\dfrac{91476}{7623}-\dfrac{693}{7623}-\dfrac{6776}{7623}+\dfrac{3267}{7623}\)
\(=\dfrac{7934}{693}\)
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)
\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)
\(=\dfrac{100}{\dfrac{-25}{11}}\)
\(=-44\)
\(E=\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(-\dfrac{41}{21}\right)}\)
\(E=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{\dfrac{213}{4}+\dfrac{187}{4}}{-\dfrac{100}{41}}\)
\(E=\dfrac{100}{-\dfrac{100}{41}}\)
\(E=-41\)
\(\dfrac{x-2}{-1,2}=\dfrac{-5}{2}\Rightarrow x=\dfrac{-5.\left(-1,2\right)}{2}+2=\dfrac{6}{2}+2=3+2=5\\ \dfrac{-6}{x+1}=\dfrac{1,8}{9}\Rightarrow x=\dfrac{-6.9}{1,8}-1=\dfrac{-54}{1,8}-1=-30-1=-31\\ \dfrac{-3}{x}=\dfrac{x}{-12}\Rightarrow x=\sqrt{\left(-12\right).\left(-3\right)}=\sqrt{36}=\sqrt{\left(\pm6\right)^2}=\pm6\)
\(\dfrac{x-4}{x-1}=\dfrac{3}{5}\\ \Rightarrow5\left(x-4\right)=3\left(x-1\right)\\ \Leftrightarrow5x-20=3x-3\\ \Leftrightarrow5x-3x=-3+20\\ \Leftrightarrow2x=17\\ \Leftrightarrow x=\dfrac{17}{2}\\ ---\\ \dfrac{1,12}{-10}=\dfrac{11,2}{x}\Rightarrow x=\dfrac{11,2.\left(-10\right)}{1,12}=\dfrac{10.1,12.\left(-10\right)}{1,12}=-100\)
a, \(A=\dfrac{10^{15}+1}{10^6+1}>1\);\(B=\dfrac{10^6+1}{10^{17}+1}< 1\)
⇒\(A>B\)
b, \(D=\dfrac{2^{2007}+3}{2^{2006}-1}=\dfrac{2^{2008}+6}{2^{2007}-2}\)
Ta có : \(\dfrac{2^{2008}-3}{2^{2007}-1}< \dfrac{2^{2008}-3}{2^{2007}-2}< \dfrac{2^{2008}+6}{2^{2007}-2}\)
⇒ \(C< D\)
c, \(M=\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)
\(N=\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)
Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\)
⇒ \(M< N\)
a: \(\left(0.5\right)^3\cdot2^3=1\)
b: \(\left(0.25\right)^2\cdot16=1\)
c: \(\left(\dfrac{3}{5}\right)^3:\left(-\dfrac{27}{1000}\right)=\dfrac{3^3}{5^3}\cdot\dfrac{-1000}{27}=\dfrac{-1000}{125}=-8\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
\(D=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}.\dfrac{6}{28}=\dfrac{5}{14}\)
\(E=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{24.25}=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)=\dfrac{2.23}{50}=\dfrac{23}{25}.\)
\(\dfrac{D}{E}=\dfrac{5}{24}.\dfrac{25}{23}=\dfrac{125}{552}.\)
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