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a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
a, \(\frac{x}{12}=-\frac{1}{24}-\frac{1}{8}\Leftrightarrow\frac{x}{12}=-\frac{1}{6}\Leftrightarrow6x=-12\Leftrightarrow x=-2\)
b, \(\frac{2}{3}x=\frac{1}{2}x+\frac{15}{12}\Leftrightarrow\frac{2x}{3}=\frac{x}{2}+\frac{15}{12}\Leftrightarrow\frac{4x}{6}-\frac{3x}{6}=\frac{15}{12}\Leftrightarrow\frac{x}{6}=\frac{15}{12}\Leftrightarrow x=\frac{15}{2}\)
c, \(\left|\frac{3}{4}-2x\right|+\frac{1}{6}=\frac{9}{2}\Leftrightarrow\left|\frac{3}{4}-2x\right|=\frac{13}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}-2x=\frac{13}{3}\\\frac{3}{4}-2x=-\frac{13}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-\frac{43}{12}\\2x=\frac{61}{12}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{43}{24}\\x=\frac{61}{24}\end{cases}}}\)
a) Ta có: \(D=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{96}\)
\(=\dfrac{63}{96}=\dfrac{21}{32}\)
b)
Sửa đề: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
Ta có: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{2048}-\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{E}{2}=\dfrac{1}{2}-\dfrac{1}{4096}=\dfrac{2047}{4096}\)
hay \(E=\dfrac{2047}{2048}\)
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)
`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)
`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)
`b)`
\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)
`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)
`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)
`c)`
\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)
`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)
a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{1}{3}\)
\(=\dfrac{-1}{6}\)
b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)
\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}+\dfrac{-7}{12}\)
\(=\dfrac{16}{9}\)
c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)
\(=\dfrac{-3}{4}+\dfrac{6}{5}\)
\(=\dfrac{9}{20}\)
\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)
\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)
Thay vào ta có :
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)
\(=1-\frac{1}{32}\)
\(=\frac{31}{32}\)
\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
Ta có :
\(\frac{1}{2}=1-\frac{1}{2}\)
\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)
...................
\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
Thay vào ta có :
\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)
\(=1-\frac{1}{256}\)
\(=\frac{255}{256}\)
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)