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a: Ta có: \(\left(x^2-2\right)\left(x^2-4\right)< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2>0\\x^2-4< 0\end{matrix}\right.\Leftrightarrow2< x^2< 4\)
mà x là số nguyên
nên \(x\in\varnothing\)
b: \(\left(x^2-2\right)\left(x^2+5\right)\left(x^2-18\right)< 0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2-18\right)< 0\)
\(\Leftrightarrow2< x^2< 18\)
mà x là số nguyên
nên \(x\in\left\{2;-2;3;-3;4;-4\right\}\)
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
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 \(x\in\left\{0;\pm1;\pm2\right\}\)
\(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(\dfrac{16}{5}:\dfrac{16}{5}+\dfrac{9}{2}\cdot\dfrac{76}{45}\right):\dfrac{-43}{2}\)
\(\Leftrightarrow-5< x< \left(1+\dfrac{38}{5}\right)\cdot\dfrac{-2}{43}=\dfrac{43}{5}\cdot\dfrac{-2}{43}=\dfrac{-2}{5}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)
=>2x-10=x+2
=>x=12
b: \(\Leftrightarrow\left(x+2\right)^2=100\)
=>x+2=10 hoặc x+2=-10
=>x=-12 hoặc x=8
c: \(\Leftrightarrow\left(2x-5\right)^3=27\)
=>2x-5=3
=>2x=8
=>x=4
a, \(\left|x+\frac{1}{3}\right|=0\Leftrightarrow x=-\frac{1}{3}\)
b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)
c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí
Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)
=> Ko có x thỏa mãn
\(|x+\frac{1}{3}|=0\)
\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)
\(|x+\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
Bài 3: Tìm x:
a. \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
=> 2x - 1 = 3
=> 2x = 4
=> x = 2
b. \(\left(x-2\right)^2=1\)
\(\Rightarrow\) \(\left(x-2\right)^2=1^2\)
=> x - 2 = 1
=> x = 3
c. \(x^{2000}=x\)
=> x = 1
d. \(\left(4x-3\right)^3=-125\)
\(\Rightarrow\left(4x-3\right)^3=\left(-5\right)^3\)
=> 4x - 3 = -5
=> 4x = -2
=> x = \(\dfrac{-1}{2}\)