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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 \(x\in\left\{0;\pm1;\pm2\right\}\)
1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)
1) \(\frac{x-1}{3}=\frac{5-x}{7}\Leftrightarrow7.\left(x-1\right)=3.\left(5-x\right)\)
\(\Leftrightarrow7x-7=15-3x\)
\(\Leftrightarrow7x+3x=15+7\)
\(\Leftrightarrow10x=22\)
\(\Leftrightarrow x=\frac{11}{5}\)
2) \(\frac{x-1}{-5}=\frac{-20}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=\left(-20\right).\left(-5\right)=100\)
\(\Leftrightarrow100=\orbr{\begin{cases}10^2\\\left(-10\right)^2\end{cases}}\)
Nếu x - 1 = 10 => x = 11
Nếu x - 1 = -10 => x = -9
Vậy ....
3) \(3\sqrt{x-3}+5=\left|-8\right|\)
\(\Leftrightarrow3\sqrt{x-3}+5=8\)
\(\Leftrightarrow3\sqrt{x-3}=3\)
\(\Leftrightarrow\sqrt{x-3}=1\) (ĐK: \(x\ge3\))
\(\Leftrightarrow\left(\sqrt{x-3}\right)^2=1^2\)
\(\Leftrightarrow x-3=1\)
\(\Leftrightarrow x=4\) (nhận)
Vậy x = 4
b) \(\left|5x-3\right|-x=7\)
\(\Rightarrow\left|5x-3\right|=7+x\)
\(\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-\left(7+x\right)\end{cases}\Rightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}\Rightarrow}\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{2}{3}\end{cases}}}\)
Vậy ....................
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~ Học tốt ~
\(a,7\left(x-3\right)=5\left(x+5\right)\)
\(\Leftrightarrow2x=46\Leftrightarrow x=23\)
\(b,\left(x^2+2x-3\right)=x^2-4\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
a ĐKXĐ x khác -5
ta có 7(x-3)=5(x+5)
7x-21=5x+5
=> 2x=26
=> x=13
b, ĐkxĐ x khác -2 x khác -3
ta có :(x-1)(x+3)=(x-2)(x+2)
x^2+2x-3-x^2 +2 = 0
=>2x+1=0
=>x=1/2
