\(\dfrac{16}{3}\) - (x + \(\dfrac{1}{2}\))2 = \(\dfrac{13}{3}\)
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VH
16 tháng 6 2019
câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)
b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)
⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)
⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)
⇒ \(x=13\)
c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)
⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)
⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)
⇒ \(x=10\)
d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
17 tháng 7 2021
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
17 tháng 7 2021
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
LD
26 tháng 7 2018
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{264}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
LD
1 tháng 1 2018
Sửa đề
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}\right)}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot3+\dfrac{5}{8}=\dfrac{3}{2}+\dfrac{5}{8}=\dfrac{17}{8}\)
AN
1 tháng 1 2018
A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{\dfrac{4}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{4^3}-\dfrac{1}{16^2})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{4^2}-\dfrac{1}{16^2})}{4-\dfrac{1}{4^3}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{16^2})}{4.-\dfrac{1}{4^2}}+\dfrac{5}{8}\)
27 tháng 8 2022
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}=\dfrac{3}{8}\)
25 tháng 3 2017
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{264}\right)}{\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{264}}\)
\(=\dfrac{1}{2}.3=\dfrac{3}{2}\)
22 tháng 7 2021
Ta có: \(B=\dfrac{\dfrac{1}{22}-\dfrac{1}{2}+\dfrac{1}{13}}{\dfrac{3}{22}-\dfrac{3}{2}+\dfrac{3}{13}}\cdot\dfrac{\dfrac{3}{4}-0.375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0.875}+\dfrac{3}{4}\)
\(=\dfrac{1}{3}\cdot\dfrac{-15}{4}+\dfrac{3}{4}\)
\(=\dfrac{-5}{4}+\dfrac{3}{4}=\dfrac{-1}{2}\)
13 tháng 6 2023
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
`16/3-(x+1/2)^2 =13/3`
`=>(x+1/2)^2=16/3-13/3`
`=>(x+1/2)^2=3/3`
`=>(x+1/2)^2=1`
`=>x+1/2=1` hoặc `x+1/2=-1`
`=>x=1-1/2` hoặc `x=-1-1/2`
`=>x=2/2-1/2` hoặc `x=-2/2-1/2`
`=>x=1/2` hoặc `x=-3/2`
16/3−(x+1/2)2=13/3
=>(�+1/2)2=16/3−13/3=>(x+1/2)2=16/3−13/3
=>(�+1/2)2=3/3=>(x+1/2)2=3/3
=>(�+1/2)2=1=>(x+1/2)2=1
=>�+1/2=1=>x+1/2=1 hoặc �+1/2=−1x+1/2=−1
=>�=1−1/2=>x=1−1/2 hoặc �=−1−1/2x=−1−1/2
=>�=2/2−1/2=>x=2/2−1/2 hoặc �=−2/2−1/2x=−2/2−1/2
=>�=1/2=>x=1/2 hoặc �=−3/2x=−3/2