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a) Giải
So sánh từng số hạng của A với B, ta thấy:
\(\dfrac{19}{41}< \dfrac{21}{41};\dfrac{23}{53}< \dfrac{23}{49}\) và \(\dfrac{29}{61}< \dfrac{33}{65}\) (vì 29.65 < 33.61)
\(\Rightarrow\dfrac{19}{41}+\dfrac{23}{53}+\dfrac{29}{61}< \dfrac{21}{41}+\dfrac{23}{49}+\dfrac{33}{65}\)
\(\Rightarrow A< B\)
Vậy A < B
b) Giải
Ta có: \(C=\dfrac{19^{20}+5}{19^{20}-8}=\dfrac{19^{20}-8+13}{19^{20}-8}=1+\dfrac{13}{19^{20}-8}\)
\(D=\dfrac{19^{21}+6}{19^{21}-7}=\dfrac{19^{21}-7+13}{19^{21}-7}=1+\dfrac{13}{19^{21}-7}\)
Vì \(19^{20}-8< 19^{21}-7\) và \(13>0\)
\(\Rightarrow\dfrac{13}{19^{20}-8}< \dfrac{13}{19^{21}-7}\)
\(\Rightarrow1+\dfrac{13}{19^{20}-8}< 1+\dfrac{13}{19^{21}-7}\)
\(\Rightarrow\) \(C< D\)
Vậy C < D.
a) \(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)
\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^2=1^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)
b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)
\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)
\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=93-3=90\)
\(\Rightarrow x=90:2=45\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
**\(\dfrac{2x-5}{6}=\dfrac{x-7}{5}\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}=\dfrac{5\left(x-7\right)}{30}\)
\(\Leftrightarrow6\left(2x-5\right)=5\left(x-7\right)\Leftrightarrow12x-30=5x-35\)
\(\Leftrightarrow12x-5x=-35+30\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)
**\(\dfrac{4x+1}{3}=\dfrac{2x-3}{4}\Leftrightarrow\dfrac{4\left(x+1\right)}{12}=\dfrac{3\left(2x-3\right)}{12}\)
\(\Leftrightarrow4\left(x+1\right)=3\left(2x-3\right)\Leftrightarrow4x+4=6x-9\)
\(\Leftrightarrow4x-6x=-9-4\Leftrightarrow-2x=-13\Leftrightarrow x=\dfrac{13}{2}\)
a, \(2-\dfrac{14}{x}=\dfrac{-22}{3}\)
\(\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)
\(\dfrac{14}{x}=\dfrac{28}{3}\)
=> \(x.28=14.3\)
\(x.28=42\)
\(x=42:28\)
\(x=\dfrac{3}{2}=1,5\)
b, \(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)
\(\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)=-\dfrac{1}{5}\)
\(\dfrac{2x}{5}=-\dfrac{1}{5}-1=\dfrac{-6}{5}\)
\(\dfrac{2x}{5}=\dfrac{-6}{5}\)
=> \(2x=-6\)
\(x=-6:2=-3\)
a)
\(2-\dfrac{14}{x}=-\dfrac{22}{3}\)
\(\Rightarrow\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)
\(\Rightarrow x=\dfrac{14.3}{28}=\dfrac{3}{2}=1,5\)
b)
\(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)
\(\Rightarrow\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)\)
\(\Rightarrow\dfrac{2x}{5}+1=-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{2x}{5}=-\dfrac{1}{5}-1=-\dfrac{6}{5}\)
Hay \(\dfrac{2x}{5}=-\dfrac{6}{5}\)
\(\Rightarrow2x=-6\)
\(\Rightarrow x=-\dfrac{6}{2}=-3\)
Chúc bạn học tốt!
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
Giải:
a) \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{-1}{2}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{-1}{2}+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{6}:\dfrac{3}{5}\)
\(\Leftrightarrow x=\dfrac{5}{18}\)
Vậy \(x=\dfrac{5}{18}\).
b) \(\left(\dfrac{1}{2}-x\right).\dfrac{2}{3}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{2}-x=\dfrac{1}{8}:\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{2}-x=\dfrac{3}{16}\)
\(\Leftrightarrow x=\dfrac{1}{2}-\dfrac{3}{16}\)
\(\Leftrightarrow x=\dfrac{5}{16}\)
Vậy \(x=\dfrac{5}{16}\).
c) \(\left|2x-\dfrac{3}{7}\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow\left|2x-\dfrac{3}{7}\right|=\dfrac{3}{4}+\dfrac{1}{2}\)
\(\Leftrightarrow\left|2x-\dfrac{3}{7}\right|=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{7}=\dfrac{5}{4}\\2x-\dfrac{3}{7}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{47}{28}\\2x=-\dfrac{23}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{47}{56}\\x=-\dfrac{23}{56}\end{matrix}\right.\)
Vậy \(x=\dfrac{47}{56}\) hoặc \(x=-\dfrac{23}{56}\).
d) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(\Leftrightarrow2\left(2x+1\right)=3\left(x-5\right)\)
\(\Leftrightarrow4x+2=3x-15\)
\(\Leftrightarrow4x-3x=-15-2\)
\(\Leftrightarrow x=-17\)
Vậy \(x=-17\).
Chúc bạn học tốt!!!
a. \(\dfrac{3}{5}x-\dfrac{2}{3}=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{5}{18}\)
b) \(\left(\dfrac{1}{2}-x\right).\dfrac{2}{3}=\dfrac{1}{8}\)
\(\Rightarrow x=\dfrac{5}{16}\)
c) \(\left|2x-\dfrac{3}{7}\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{7}\right|=\dfrac{5}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{7}=\dfrac{5}{4}\\2x-\dfrac{3}{7}=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{47}{56}\\x=\dfrac{-23}{56}\end{matrix}\right.\)
d) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(\Rightarrow4x+2=3x-15\)
\(\Rightarrow x=-17\).
\(2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)
\(2x+\left(1+\dfrac{1}{6}\right)+\left(1+\dfrac{1}{12}\right)+\left(1+\dfrac{1}{20}\right)+\left(1+\dfrac{1}{30}\right)+\left(1+\dfrac{1}{42}\right)+\left(1+\dfrac{1}{56}\right)+\left(1+\dfrac{1}{72}\right)+\left(1+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)
~ Chúc bạn học tốt ~
2x+\(\dfrac{7}{2.3}+\dfrac{13}{3.4}+\dfrac{21}{4.5}+...+\dfrac{91}{9.10}=10\)
\(2x+7\left(\dfrac{1}{2.3}\right)+13\left(\dfrac{1}{3.4}\right)+...+91\left(\dfrac{1}{9.10}\right)=10\)
\(2x+\left(7+13+...+91\right)\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)=10\)
\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)
\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)
\(2x+336.\dfrac{2}{5}=10\)
\(2x+\dfrac{672}{5}=10\)
\(2x=10-\dfrac{672}{5}\)
\(2x=-\dfrac{622}{5}\)
\(x=-\dfrac{311}{5}\)
Tick nha 0o0^^^Nhi^^^0o0
\(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)
\(\Rightarrow\dfrac{2x+19}{21}-\dfrac{2x+17}{23}-\dfrac{2x+7}{33}+\dfrac{2x+5}{35}=0\)
\(\Rightarrow\left(\dfrac{2x+19}{21}+1\right)-\left(\dfrac{2x+17}{23}+1\right)-\left(\dfrac{2x+7}{33}+1\right)+\left(\dfrac{2x+5}{35}+1\right)=0\)
\(\Rightarrow\dfrac{2x+40}{21}-\dfrac{2x+40}{23}-\dfrac{2x+40}{33}+\dfrac{2x+40}{35}=0\)
\(\Rightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)
\(\Rightarrow2x+40=0\Rightarrow x=-20\)( do \(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}>0\))
Ta có: \(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)
\(\Leftrightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow2x+40=0\)
hay x=-20