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a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)
=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)
=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)
b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)
c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)
d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)
=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)
=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)
e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)
=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)
=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)
f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)
=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)
=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)
=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)
\(\left(3x-1\right)^3=64\)
\(\left(3x-1\right)^3=4^3\)
3x - 1 = 4
3x = 4 + 1
3x = 5
x = 5/3
\(\left(3x-1\right)^3=64\)
\(\Leftrightarrow3x-1=4\)
\(\Leftrightarrow3x=5\Leftrightarrow x=\frac{5}{3}\)
2. 2(x-1) +3 2-x) =- 1
\(\Leftrightarrow2x-2+6x-3=-1\)
\(\Leftrightarrow8x-5=-1\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
3. ( n2 + 3 ) chia hết cho ( n - 1)
\(\Leftrightarrow n^2-1+4⋮n-1\Leftrightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\)
Vì n thuộc Z => ( n-1) ( n+1) thuộc Z
\(\Rightarrow\left(n-1\right)\left(n+1\right)+4⋮n-1\Leftrightarrow4⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Phần còn lại bn tự làm
a) \(\left(\frac{2x}{3}-3\right)\div\left(-10\right)=\frac{2}{5}\)
\(\Rightarrow\left(\frac{2x}{3}-\frac{9}{3}\right).-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow\frac{2x-9}{3}.-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow\frac{9-2x}{30}=\frac{2}{5}\Rightarrow\left(9-2x\right).5=60\)
\(\Rightarrow45-10x=60\Rightarrow-10x=15\Rightarrow x=-\frac{15}{10}=-1,5\)
b)\(\left(x+5\right)^3=64\)
\(\Rightarrow\left(x+5\right)^3=4^3\)
\(\Rightarrow x+5=4\Rightarrow x=-1\)
A, 2x/3-3=-4
2x/3=-1
2x=-3
X=-3/2
B,(x+5)^3=4^3
=> x+5=4
X=-1
k nha
Ta có:(x+1)^3=64
=>(x+1)^3=4^3
=>x+1=4
=>x=4-1=3
Vậy x=3
Tick cho mình nhé
\(1+\left(x-\frac{2}{3}\right)^3=\frac{37}{64}\)
\(\left(x-\frac{2}{3}\right)^3=\frac{37}{64}-1\)
\(\left(x-\frac{2}{3}\right)^3=\frac{-27}{64}\)
\(\left(x-\frac{2}{3}\right)^3=\left(\frac{-3}{4}\right)^3\)
\(\Rightarrow x-\frac{2}{3}=\frac{-3}{4}\)
\(\Rightarrow x=\frac{-3}{4}+\frac{2}{3}\)
\(\Rightarrow x=\frac{-1}{12}\)
\(1+\left(x-\frac{2}{3}\right)^3=\frac{37}{64}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{37}{64}-1\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{-27}{64}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\left(\frac{-3}{4}\right)^3\)
\(\Rightarrow x-\frac{2}{3}=\frac{-3}{4}\)
\(\Rightarrow x=\frac{-3}{4}+\frac{2}{3}\)
\(\Rightarrow x=\frac{-9}{12}+\frac{8}{12}\)
\(\Rightarrow\frac{-1}{12}\)