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Bài 4
d. 450 : [ 41 - (2x - 5) ] = 32 . 5
450 : [ 41 - (2x - 5) ] = 9 . 5
450 : [ 41 - (2x - 5) ] = 45
[ 41 - (2x - 5) ] = 450 : 45
41 - (2x - 5) = 10
(2x - 5) = 41 - 10
2x - 5 = 31
2x = 31 + 5
2x = 36
x = 36 : 2
x = 1
e. 30 : (x - 7) = 1519 : 158
30 : (x - 7) = 15
x - 7 = 30 : 15
x - 7 = 2
x = 2 + 7
x = 9
f. (2x - 3)3 = 125
2x - 3 = 5
2x = 5 + 3
2x = 8
x = 8 : 2
x = 4
tk cho cj nha
\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)
\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)
\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)
\(\Rightarrow x=3\)
\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Leftrightarrow25-5x=6x-8\)
\(\Leftrightarrow-5x-6x=-8-25\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3
i) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
\(\Rightarrow\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)
\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x=\dfrac{4}{11}-\dfrac{33}{182}\)
\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)
\(\Rightarrow x=\dfrac{89}{17}:\dfrac{365}{2002}=\dfrac{178178}{6205}\)
j) \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}-\left(-\dfrac{7}{4}\right)=\dfrac{41}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=-\dfrac{41}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=11\Rightarrow x=\dfrac{11}{2}\\2x=-\dfrac{19}{2}\Rightarrow x=-\dfrac{19}{4}\end{matrix}\right.\)
k) \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)\(=\left(-\dfrac{3}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\Rightarrow x=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\Rightarrow x=-\dfrac{4}{5}\end{matrix}\right.\)
l) \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-32}{27}-\left(-\dfrac{24}{27}\right)=-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Rightarrow3x=-\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{1}{27}\)
j, \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}\)
\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}-\dfrac{17}{2}\)
\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-41}{4}\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=\dfrac{-41}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-19}{4}\end{matrix}\right.\)
k, \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow x+\dfrac{1}{5}=\pm\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=\dfrac{-3}{5}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)
l, \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-24}{27}\)
\(\Rightarrow-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-19}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{19}{27}\)
\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{\sqrt[3]{19}}{3}\)
\(\Rightarrow3x=\dfrac{\sqrt[3]{19}}{3}+\dfrac{7}{19}\)
\(\Rightarrow...\)
Ta có: \(\dfrac{1}{2}\cdot y+\dfrac{2}{3}\cdot y=\dfrac{7}{6}\Rightarrow y\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{7}{6}\Rightarrow\dfrac{7}{6}y=\dfrac{7}{6}\Rightarrow y=\dfrac{7}{6}:\dfrac{7}{6}=1\)
Vậy \(D=\left\{1\right\}\)
Bài 6:
Số kẹo chia đều vào các đĩa nên số đĩa là ước của \(28\).
Có \(Ư\left(28\right)=\left\{1,2,4,7,14,28\right\}\)mà số đĩa lớn hơn \(5\)và nhỏ hơn \(15\)
nên có hai cách chia là \(7\)đĩa và \(14\)đĩa.
Với cách chia \(7\)đĩa mỗi đĩa có số kẹo là \(28\div7=4\)chiếc.
Với cách chia \(14\)đĩa mỗi đĩa có số kẹo là \(28\div14=2\)chiếc.
Bài 16:
\(e,\left(6n+11\right)⋮\left(2n+3\right)\\ \Rightarrow\left[3\left(2n+3\right)+2\right]⋮\left(2n+3\right)\\ \Rightarrow2⋮\left(2n+3\right)\\ \Rightarrow2n+3\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Rightarrow n\in\varnothing\left(n\in N\right)\\ g,\left(3n+5\right)⋮\left(2n+1\right)\\ \Rightarrow\left(6n+10\right)⋮\left(2n+1\right)\\ \Rightarrow\left[3\left(2n+1\right)+7\right]⋮\left(2n+1\right)\\ \Rightarrow7⋮\left(2n+1\right)\\ \Rightarrow2n+1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Rightarrow n\in\left\{0;3\right\}\left(n\in N\right)\)