`#040911`
`b)`
`2/3 \div [1/3 - ( (-2)/5 )] + 3/11 \div ( (-2)^2 )/121`
`= 2/3 \div 11/15 + 3/11 \div 4/121`
`= 10/11 + 33/4`
`= 403/44`
`c)`
`5/7 \div (5/22 - 1/11) + 5/7 \div (1 - 5/8) + ( (-13)/15 )^0`
`= 5/7 \div 3/22 + 5/7 \div 3/8 + 1`
`= 5/7 * 22/3 + 5/7 * 8/3 + 1`
`= 5/7 * (22/3 + 8/3) + 1`
`= 5/7 * 30/3 + 1`
`= 5/7 * 10 + 1`
`= 50/7 + 1`
`= 57/7`
___
`b)`
`(2x + 1)^2 = 25/49`
\(\Rightarrow\left(2x+1\right)^2=\left(\pm\dfrac{5}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{5}{7}\\2x+1=-\dfrac{5}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{5}{7}-1\\2x=-\dfrac{5}{7}-1\end{matrix}\right.\\ \Rightarrow \left[{}\begin{matrix}2x=-\dfrac{2}{7}\\2x=-\dfrac{12}{7}\end{matrix}\right.\\ \Rightarrow \left[{}\begin{matrix}x=-\dfrac{1}{7}\\x=-\dfrac{6}{7}\end{matrix}\right.\)
Vậy, `x \in {-1/7; -6/7}.`
`c)`
`( 6 1/2 - (x + 1) ) \div (-5)/4 = 2/5`
`=> (6 1/2 - x - 1) * (-4)/5 = 2/5`
`=> (11/2 - x) = 2/5 \div (-4)/5`
`=> 11/2 - x = -1/2`
`=> x = 11/2 - (-1/2)`
`=> x = 6`
Vậy, `x = 6.`
\(b2,\left(2x+1\right)^2=\dfrac{25}{49}\\\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=\left(\dfrac{5}{7}\right)^2\\\left(2x+1\right)^2=\left(-\dfrac{5}{7}\right)^2\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{5}{7}\\2x+1=-\dfrac{5}{7}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{7}-1=-\dfrac{2}{7}\\2x=-\dfrac{5}{7}-1=-\dfrac{12}{7}\end{matrix}\right. \\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{7}:2=-\dfrac{1}{7}\\x=-\dfrac{12}{7}:2=-\dfrac{6}{7}\end{matrix}\right.\)
\(c2,\\ \left(6\dfrac{1}{2}-\left(x+1\right)\right):\dfrac{-5}{4}=\dfrac{2}{5}\\ \Leftrightarrow\left(\dfrac{13}{2}-1-x\right).\dfrac{-4}{5}=\dfrac{2}{5}\\ \Leftrightarrow\left(\dfrac{11}{2}-x\right)=\dfrac{2}{5}:\dfrac{-4}{5}=\dfrac{2}{5}.\dfrac{5}{-4}\\ \Leftrightarrow\left(\dfrac{11}{2}-x\right)=-\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{11}{2}-\left(-\dfrac{1}{2}\right)\\ \Leftrightarrow x=\dfrac{12}{2}=6\)
\(b1,\\ \dfrac{2}{3}.\left[\dfrac{1}{3}-\left(-\dfrac{2}{5}\right)\right]+\dfrac{3}{11}:\dfrac{\left(-2\right)^2}{121}\\ =\dfrac{2}{3}.\left(\dfrac{1}{3}+\dfrac{2}{5}\right)+\dfrac{3}{11}:\dfrac{4}{121}\\ =\dfrac{2}{3}.\dfrac{11}{15}+\dfrac{3}{11}.\dfrac{121}{4}\\ =\dfrac{22}{45}+\dfrac{3.11.11}{11.4}\\ =\dfrac{22}{45}+\dfrac{33}{4}\\ =\dfrac{22.4+33.45}{180}=\dfrac{1573}{180}\)
\(c1,\dfrac{5}{7}:\left(\dfrac{5}{22}-\dfrac{1}{11}\right)+\dfrac{5}{7}:\left(1-\dfrac{5}{8}\right)+\left(-\dfrac{13}{15}\right)^0\\ =\dfrac{5}{7}.\left(\dfrac{5-2}{22}\right)+\dfrac{5}{7}:\dfrac{3}{8}+1\\ =\dfrac{5}{7}.\dfrac{22}{3}+\dfrac{5}{7}.\dfrac{8}{3}+1\\ =\dfrac{110}{21}+\dfrac{40}{21}+1\\ =\dfrac{150}{21}+1\\ =7\dfrac{3}{21}+1\\ =7\dfrac{1}{7}+1=8\dfrac{1}{7}\)
