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a)Ta có: 1/2-(1/3+1/4)= -1/12
1/48-(1/16-1/6)=1/8
suy ra: -1/12<x<1/8
<=> -2/24<x<3/24
=>x thuộc:(-1/24 ;0 ;1/24 ;2/24 ;3/24)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< x< \dfrac{1}{48}+\dfrac{5}{48}=\dfrac{6}{48}=\dfrac{1}{8}\)
\(\Leftrightarrow-\dfrac{1}{12}< x< \dfrac{1}{8}\)
=>x=0
c: \(\Leftrightarrow x=\dfrac{-1}{2}\cdot\dfrac{1}{4}=\dfrac{-1}{8}\)
d: \(\Leftrightarrow x^8=x^7\)
=>x(x-1)=0
=>x=0(loại) hoặc x=1(nhận)
e: \(\Leftrightarrow3^x=\dfrac{3^{10}}{3^9}=3\)
hay x=1
f: =>x-1=20
hay x=21
\(e,\frac{22}{15}-x=-\frac{8}{27}\)
=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)
=> \(x=\frac{22}{15}+\frac{8}{27}\)
=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)
\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)
=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)
=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)
=> \(\frac{2x-5}{5}=-\frac{5}{4}\)
=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)
=> \(2x=-\frac{5}{4}\)
=> \(x=-\frac{5}{8}\)
\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)
=> \(-\frac{9}{4}x+\frac{37}{4}=20\)
=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)
=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)
\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)
=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)
=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)
=> \(-10\le x\le-\frac{13}{7}\)
Đến đây tìm x
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Rightarrow\frac{-1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(\Rightarrow\frac{-1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Rightarrow-1\le x< 7\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5;6\right\}\)
= 9/12 - 10/12 =< x/12 < 1 -( 8/12 - 3/12)
= -1/12 =< x/12 < 7/12
=> x thuộc -1,0,1,2,3,4,5,6
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\frac{-1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(\frac{-1}{12}\le x< \frac{7}{12}\)
mà \(x\in z\)
=> x = 0
KL: x = 0
\(=-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5;6\right\}\)
vậy : ................
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)(1)
VT:\(\frac{3}{4}-\frac{5}{6}=-\frac{1}{12}\)
VP:\(1-\left(\frac{2}{3}-\frac{1}{4}\right)=\frac{7}{12}\)
(1)\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Leftrightarrow-1\le x< 7\)
mà \(x\inℤ\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5;6\right\}\)