Bài 2:
\(a,\dfrac{3}{4}:\left(x+\dfrac{1}{2}\right)-\dfrac{5}{6}=-\left(\dfrac{1}{2}\right)^2\\
=>\dfrac{3}{4}:\left(x+\dfrac{1}{2}\right)=-\dfrac{1}{4}+\dfrac{5}{6}\\
=>\dfrac{3}{4}:\left(x+\dfrac{1}{2}\right)=\dfrac{7}{12}\\
=>x+\dfrac{1}{2}=\dfrac{3}{4}:\dfrac{7}{12}\\
=>x+\dfrac{1}{2}=\dfrac{9}{7}\\
=>x=\dfrac{9}{7}-\dfrac{1}{2}\\
=>x=\dfrac{11}{14}\\
b,\left(2x-3\right)^2-\dfrac{1}{4}=6^{20}:6^{19}\\
=>\left(2x-3\right)^2-\dfrac{1}{4}=6\\
=>\left(2x-3\right)^2=6+\dfrac{1}{4}=\dfrac{25}{4}=\left(\dfrac{5}{2}\right)^2\\
TH1:2x-3=\dfrac{5}{2}\\
=>2x=3+\dfrac{5}{2}=\dfrac{11}{2}\\
=>x=\dfrac{11}{2}:2=\dfrac{11}{4}\\
TH2:2x-3=\dfrac{-5}{2}\\
=>2x=3-\dfrac{5}{2}=\dfrac{1}{2}\\
=>x=\dfrac{1}{2}:2=\dfrac{1}{4}\)
\(c,\left(-\dfrac{1}{3}\right)^{x-3}=\dfrac{1}{81}\\ =>\left(-\dfrac{1}{3}\right)^{x-3}=\left(\dfrac{1}{3}\right)^4\\ =>\left(-\dfrac{1}{3}\right)^{x-3}=\left(-\dfrac{1}{3}\right)^4\\ =>x-3=4\\ =>x=3+4=7\\ d,\dfrac{x+1}{2022}+\dfrac{x+3}{2020}+\dfrac{x+5}{2018}+\dfrac{x+7}{2016}+4=0\\ =>\left(\dfrac{x+1}{2022}+1\right)+\left(\dfrac{x+3}{2020}+1\right)+\left(\dfrac{x+5}{2018}+1\right)+\left(\dfrac{x+7}{2016}+1\right)=0\\ =>\dfrac{x+2023}{2022}+\dfrac{x+2023}{2020}+\dfrac{x+2023}{2018}+\dfrac{x+2023}{2016}=0\\ =>\left(x+2023\right)\left(\dfrac{1}{2022}+\dfrac{1}{2020}+\dfrac{1}{2018}+\dfrac{1}{2016}\right)=0\\ =>x+2023=0\\ =>x=-2023\)
Bài 1:
a: \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2025^0\)
\(=\dfrac{9}{2}\cdot4+\dfrac{3}{4}\cdot1\)
\(=18+\dfrac{3}{4}=\dfrac{75}{4}\)
b: \(\dfrac{21}{11}-3\dfrac{4}{9}:\dfrac{24}{5}-\dfrac{5}{9}:\dfrac{24}{5}\)
\(=\dfrac{21}{11}-\dfrac{31}{9}\cdot\dfrac{5}{24}-\dfrac{5}{9}\cdot\dfrac{5}{24}\)
\(=\dfrac{21}{11}-\dfrac{5}{24}\cdot\left(\dfrac{31}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{21}{11}-\dfrac{5}{24}\cdot4=\dfrac{21}{11}-\dfrac{5}{6}=\dfrac{126}{66}-\dfrac{55}{66}=\dfrac{71}{66}\)
c: \(\dfrac{15^5}{5^5}-\left(-0,25\right)^2\cdot4^2\)
\(=3^5-\left(0,25\cdot4\right)^2\)
=243-1
=242
d: \(\dfrac{3^{12}\cdot9^{12}+3^{14}\cdot9^{13}}{3^{15}\cdot9^{12}+8\cdot3^{14}\cdot9^{12}}\)
\(=\dfrac{3^{12}\cdot3^{24}+3^{14}\cdot3^{26}}{3^{15}\cdot3^{24}+8\cdot3^{14}\cdot3^{24}}=\dfrac{3^{38}+3^{40}}{3^{39}+8\cdot3^{38}}\)
\(=\dfrac{3^{38}\left(1+3^2\right)}{3^{38}\left(1+8\right)}=\dfrac{1+9}{9}=\dfrac{10}{9}\)
Bài 2:
a: \(\dfrac{3}{4}:\left(x+\dfrac{1}{2}\right)-\dfrac{5}{6}=\left(-\dfrac{1}{2}\right)^2\)
=>\(\dfrac{3}{4}:\left(x+\dfrac{1}{2}\right)-\dfrac{5}{6}=\dfrac{1}{4}\)
=>\(\dfrac{3}{4}:\left(x+\dfrac{1}{2}\right)=\dfrac{5}{6}+\dfrac{1}{4}=\dfrac{10}{12}+\dfrac{3}{12}=\dfrac{13}{12}\)
=>\(x+\dfrac{1}{2}=\dfrac{3}{4}:\dfrac{13}{12}=\dfrac{3}{4}\cdot\dfrac{12}{13}=\dfrac{9}{13}\)
=>\(x=\dfrac{9}{13}-\dfrac{1}{2}=\dfrac{18}{26}-\dfrac{13}{26}=\dfrac{5}{26}\)
b: \(\left(2x-3\right)^2-\dfrac{1}{4}=6^{20}:6^{19}\)
=>\(\left(2x-3\right)^2=6+\dfrac{1}{4}=6,25\)
=>\(\left[{}\begin{matrix}2x-3=2,5\\2x-3=-2,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5,5\\2x=0,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,75\\x=0,25\end{matrix}\right.\)
c: \(\left(-\dfrac{1}{3}\right)^{x-3}=\dfrac{1}{81}\)
=>\(\left(-\dfrac{1}{3}\right)^{x-3}=\left(-\dfrac{1}{3}\right)^4\)
=>x-3=4
=>x=7
d: \(\dfrac{x+1}{2022}+\dfrac{x+3}{2020}+\dfrac{x+5}{2018}+\dfrac{x+7}{2016}+4=0\)
=>\(\left(\dfrac{x+1}{2022}+1\right)+\left(\dfrac{x+3}{2020}+1\right)+\left(\dfrac{x+5}{2018}+1\right)+\left(\dfrac{x+7}{2016}+1\right)=0\)
=>\(\dfrac{x+2023}{2022}+\dfrac{x+2023}{2020}+\dfrac{x+2023}{2018}+\dfrac{x+2023}{2016}=0\)
=>x+2023=0
=>x=-2023
`#3107.101107`
`2.`
`a)`
`3/4 \div (x + 1/2) - 5/6 = -(1/2)^2`
`=> 3/4 \div (x + 1/2) = -1/4 + 5/6`
`=> 3/4 \div (x + 1/2) = 7/12`
`=> x + 1/2 = 3/4 \dvi 7/12`
`=> x + 1/2 = 9/7`
`=> x = 9/7 - 1/2`
`=> x = 11/14`
Vậy, `x = 11/14`
`b)`
\(\left(-\dfrac{1}{3}\right)^{x-3}=\dfrac{1}{81}\\ \Rightarrow\left(-\dfrac{1}{3}\right)^{x-3}=\left(-\dfrac{1}{3}\right)^4\\ \Rightarrow x-3=4\\ \Rightarrow x=4+3\\ \Rightarrow x=7\)
`c)`
\(\left(2x-3\right)^2-\dfrac{1}{4}=6^{20}\div6^{19}\\ \Rightarrow\left(2x-3\right)^2-\dfrac{1}{4}=6\\ \Rightarrow\left(2x-3\right)^2=6+\dfrac{1}{4}\\ \Rightarrow\left(2x-3\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(2x-3\right)^2=\left(\pm\dfrac{5}{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{5}{2}\\2x-3=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{5}{2}+3\\2x=-\dfrac{5}{2}+3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{11}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy, `x \in {11/4; 1/4}`
`d)`
\(\dfrac{x+1}{2022}+\dfrac{x+3}{2020}+\dfrac{x+5}{2018}+\dfrac{x+7}{2016}+4=0\)
`=> ((x + 1)/2022 + 1) + ((x + 3)/2020 + 1) + ((x + 5)/2018 + 1) + ((x + 7)/2016 + 1) = 0`
`=> (x + 1 + 2022)/2022 + (x + 3 + 2020)/2020 + (x + 5 + 2018)/2018 + (x + 7 + 2016)/2016 = 0`
`=> (x + 2023)/2022 + (x + 2023)/2020 + (x + 2023)/2018 + (x + 2023)/2016 = 0`
`=> (x + 2023)(1/2022 + 1/2020 + 1/2018 + 1/2016) = 0`
Vì `1/2022 + 1/2020 + 1/2018 + 1/2016 \ne 0`
`=> x + 2023 = 0`
`=> x = -2023`
Vậy, `x = -2023.`
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