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Bài 1:
\(a)\left(x+\dfrac{2}{3}\right)^3=\dfrac{125}{64}.\\ \Leftrightarrow\left(x+\dfrac{2}{3}\right)^3=\left(\dfrac{5}{4}\right)^3.\\ \Rightarrow x+\dfrac{2}{3}=\dfrac{5}{4}.\\ \Leftrightarrow x=\dfrac{7}{12}.\)
\(b)\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{343}.\\\Leftrightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{7}\right) ^3.\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{2}{7}.\\ \Leftrightarrow x=\dfrac{11}{14}.\)
Bài 2:
\(a)\left(x-\dfrac{1}{3}\right)^2=\dfrac{25}{9}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{5}{3}\right)^2.\\\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{-5}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{5}{3}.\\x-\dfrac{1}{3}=\dfrac{-5}{3}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-4}{3}.\end{matrix}\right.\)
\(b)\left(x-\dfrac{3}{4}\right)^2=\dfrac{49}{16}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{7}{4}\right)^2.\\\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{-7}{4}\right)^2.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{4}.\\x-\dfrac{3}{4}=\dfrac{-7}{4}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}.\\x=-1.\end{matrix}\right.\)
Bài 3:
a: =>1<x<=4
hay \(x\in\left\{2;3;4\right\}\)
b: =>-104<x<-99
hay \(x\in\left\{-103;-102;-101;-100\right\}\)
c: ⇔n+2∈{1;−1;5;−5}⇔n+2∈{1;−1;5;−5}
hay n∈{−1;−3;3;−7}n∈{−1;−3;3;−7}
d: ⇔n+2∈{1;−1;2;−2;4;−4}⇔n+2∈{1;−1;2;−2;4;−4}
hay n∈{−1;−3;0;−4;2;−6}n∈{−1;−3;0;−4;2;−6}
a: ⇔n−1∈{1;−1;5;−5}⇔n−1∈{1;−1;5;−5}
hay n∈{2;0;6;−4}
b: \(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-2;1;-3;3;-5\right\}\)
c: \(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
d: \(\Leftrightarrow n+2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{-1;-3;0;-4;2;-6\right\}\)
a: \(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
Để 56a3b chia hết cho 4 => b =6
Để 56a36 chia hết cho 9 => ( 6+5+3+6+a) chia hết cho 9
=> ( 20+a) chia hết cho 9 => a =7
Số đó là 56736
❆ Để 56a3b⋮4 thì b=6
❆ Để 56a3b⋮9 thì (6+5+3+6+a)⋮9
⇒(20+a)⋮9→a=7
✔ Vậy số cần tìm là: 56736.