Giải phương trình :
a) 13-\(\sqrt{\left(8x-1\right)^2}\) =\(\sqrt{x^2}\)
b) \(\sqrt{\left(x+1^2\right)}\)+\(\sqrt{\left(2x+3\right)^2}\)=3
c) \(\sqrt{\left(-x+\dfrac{2}{3}\right)^2}\)+ \(\sqrt{\left(x-1\dfrac{2}{5}\right)^2}\)+3=0
d) 6-\(\sqrt{x^2-6x+9}\)=\(\dfrac{5}{9}\)
Giúp mình với , mình đang cần gấp !!
a) \(13-\sqrt{\left(8x-1\right)^2}=\sqrt{x^2}\) (*)
\(\Leftrightarrow13-\left|8x-1\right|=\left|x\right|\)
Th1: \(8x-1\ge0\Leftrightarrow x\ge\dfrac{1}{8}\)
(*) \(\Leftrightarrow13-8x+1=x\Leftrightarrow9x=14\Leftrightarrow x=\dfrac{14}{9}\left(N\right)\)
Th2: \(x\le0\)
(*) \(\Leftrightarrow13+8x-1=-x\Leftrightarrow9x=-12\Leftrightarrow x=-\dfrac{4}{3}\left(N\right)\)
Th3: \(\left\{{}\begin{matrix}8x-1\ge0\\x\le0\end{matrix}\right.\Leftrightarrow\dfrac{1}{8}\le x\le0\) (vô lý)
Th4: \(\left\{{}\begin{matrix}8x-1\le0\\x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le\dfrac{1}{8}\)
(*) \(\Leftrightarrow13-8x+1=x\Leftrightarrow9x=14\Leftrightarrow x=\dfrac{14}{9}\left(L\right)\)
Kl: x= 14/9 , x= -4/3
b) \(\sqrt{\left(x+1\right)^2}+\sqrt{\left(2x+3\right)^2}=3\Leftrightarrow\left|x+1\right|+\left|2x+3\right|=3\)(*)
Th1: \(x\ge-1\)
(*) \(\Leftrightarrow x+1+2x+3=3\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\left(N\right)\)
Th2: \(x\le-\dfrac{3}{2}\)
(*) \(\Leftrightarrow-x-1-2x-3=3\Leftrightarrow-3x=7\Leftrightarrow x=-\dfrac{7}{3}\left(N\right)\)
Th3: \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\le0\end{matrix}\right.\Leftrightarrow-1\le x\le-\dfrac{3}{2}\) (vô lý)
Th4: \(\left\{{}\begin{matrix}x+1\le0\\2x+3\ge0\end{matrix}\right.\Leftrightarrow-\dfrac{3}{2}\le x\le-1\)
(*) \(\Leftrightarrow-x-1-2x-3=3\Leftrightarrow-3x=7\Leftrightarrow x=-\dfrac{7}{3}\left(L\right)\)
Kl: x= -1/3 , x= -7/3
c/ \(\sqrt{\left(-x+\dfrac{2}{3}\right)^2}+\sqrt{\left(x-1\dfrac{2}{5}\right)^2}+3=0\)
\(\Leftrightarrow\left|-x+\dfrac{2}{3}\right|+\left|x-\dfrac{7}{5}\right|+3=0\)
\(\Leftrightarrow\left|-x-\dfrac{2}{3}\right|=-3-\left|x-\dfrac{2}{5}\right|\)
Ta thấy: \(VP=\left|-x-\dfrac{2}{3}\right|\ge0\forall x\)
\(VT=-3-\left|x-\dfrac{2}{5}\right|\le-3< 0\)
=> Dấu ''='' k xảy ra
Vậy pt vô nghiệm
c) \(\sqrt{\left(-x+\dfrac{2}{3}\right)^2}+\sqrt{\left(x-1\dfrac{2}{5}\right)^2}+3=0\)
\(\Leftrightarrow\left|-x+\dfrac{2}{3}\right|+\left|x-\dfrac{7}{5}\right|+3=0\) (*)
Ta có: VT >/ 3 ; VP =0
vậy ptvn
d) \(6-\sqrt{x^2-6x+9}=\dfrac{5}{9}\Leftrightarrow\left|x-3\right|=\dfrac{49}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=-\dfrac{49}{9}\\x-3=\dfrac{49}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{22}{9}\left(N\right)\\x=\dfrac{76}{9}\left(N\right)\end{matrix}\right.\)
Kl: x= -22/9, x= 76/9
d/ \(6-\sqrt{x^2-6x+9}=\dfrac{5}{9}\)
\(\Leftrightarrow\left|x-3\right|=6-\dfrac{5}{9}=\dfrac{49}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\dfrac{49}{9}\\x-3=-\dfrac{49}{9}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{76}{9}\\x=-\dfrac{22}{9}\end{matrix}\right.\)(tm)
Vậy.....
