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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
\(|x+\frac{1}{3}|=0\)
\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)
\(|x+\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
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 \(x\in\left\{0;\pm1;\pm2\right\}\)
a, \(\left(x-3\right)\left(x+2\right)>0\)
th1 : \(\hept{\begin{cases}x-3>0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x>-2\end{cases}\Rightarrow}x>3}\)
th2 : \(\hept{\begin{cases}x-3< 0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x< -3\end{cases}\Rightarrow}x< -3}\)
vậy x > 3 hoặc x < -3
b, \(\left(x+5\right)\left(x+1\right)< 0\)
th1 : \(\hept{\begin{cases}x+5>0\\x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-5\\x< -1\end{cases}\Rightarrow x\in\left\{-4;-3;-2\right\}}}\)
th2 : \(\hept{\begin{cases}x+5< 0\\x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x< -5\\x>-1\end{cases}\Rightarrow}x\in\varnothing}\)
vậy x = -4; -3; -2
c, \(\frac{x-4}{x+6}\le0\)
xét \(\frac{x-4}{x+6}=0\)
\(\Rightarrow x-4=0;x\ne-6\)
\(\Rightarrow x=4\ne-6\)
xét \(\frac{x-4}{x+5}< 0\)
th1 : \(\hept{\begin{cases}x-4< 0\\x+5>0\end{cases}\Rightarrow\hept{\begin{cases}x< 4\\x>-5\end{cases}\Rightarrow}x\in\left\{3;2;1;0;-1;-2;-3;-4\right\}}\)
th2 : \(\hept{\begin{cases}x-4>0\\x+5< 0\end{cases}\Rightarrow\hept{\begin{cases}x>4\\x< -5\end{cases}\Rightarrow x\in\varnothing}}\)
d tương tự c
\(\frac{\left(x-6\right)}{x-7}\ge0\)
Th1: x - 6 < 0
<=> x - 6 + 6 < 0 + 6
<=> x - 6 + 6 > 0 + 6
=> x < 6
Th2: x - 7
<=> x - 7 + 7 < 0 + 7
<=> x - 7 + 7 > 0 + 7
=> x > 7
=> x < 6 hoặc x > 7
a)\(\sqrt{x}=4\Leftrightarrow x=4^2\Leftrightarrow x=16\)
b)\(\sqrt{x-2}=3\Leftrightarrow x-2=3^2\Leftrightarrow x=9-2=7\)
c)\(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\Leftrightarrow\dfrac{x}{3}-\dfrac{7}{6}=\dfrac{1}{36}\Leftrightarrow\dfrac{x}{3}=-\dfrac{41}{36}\Leftrightarrow x=-\dfrac{41}{12}\)
d)\(x^2=7vớix< 0\)
\(\Leftrightarrow\left(-x\right)^2=7\Leftrightarrow-x=\sqrt{7}\Leftrightarrow x=-\sqrt{7}\)
e)\(x^2-4=0với>0\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\sqrt{4}=2\)
f)\(\left(2x+7\sqrt{7}\right)^2=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}+343=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}=-336\)
\(\Leftrightarrow4x^2=28\left(12-\sqrt{7}\right)\Leftrightarrow x^2=\dfrac{28\left(12-\sqrt{7}\right)}{4}=7\left(12-\sqrt{7}\right)\)
\(\Leftrightarrow x=\sqrt{7\left(12-\sqrt{7}\right)}=\sqrt{84-7\sqrt{7}}\)
a) Ta có: \(\hept{\begin{cases}\left|y-1\right|\ge0\forall y\\\left|5-x\right|\ge0\forall x\end{cases}\Rightarrow\left|y-1\right|+\left|5-x\right|\ge0\forall}x;y\)
Mà \(\left|y-1\right|+\left|5-x\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|y-1\right|=0\\\left|5-x\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}y-1=0\\5-x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=5\end{cases}}}\)
Vậy \(\hept{\begin{cases}y=1\\x=5\end{cases}}\)
b) Ta có: \(\left|y-6\right|\ge0\forall y\)
\(\Rightarrow\left|y-6\right|>0\Leftrightarrow y\ne6\)
\(\Rightarrow\)Để \(\frac{\left|y-6\right|}{x+2}>0\)thì \(\hept{\begin{cases}y\ne6\\x+2>0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}y\ne6\\x>-2\end{cases}}\)
Vậy \(\hept{\begin{cases}y\ne6\\x>-2\end{cases}}\)
c) Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2>0\Leftrightarrow x\ne0\)
Để \(\frac{x^2-1}{x^2}>0\Leftrightarrow\hept{\begin{cases}x^2-1>0\\x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x>1\\x\ne0\end{cases}\Leftrightarrow}x>1}\)
Vậy \(x>1\)
Tham khảo nhé~
Do đó, x = 0 hoặc x = 6 hoặc x = -6
Chọn đáp án C