Question

giải pt

\(\sqrt{9x^2}=2x+1\)

\(\sqrt{x^4}=7\)

\(\sqrt{x^2}=7\)

\(\sqrt{x^2}=\left|-8\right|\)

\(\sqrt{x^4}=9\)

Answer 1

\(\sqrt{9x^2}=2x+1\\ \Leftrightarrow\sqrt{\left(3x\right)^2}=2x+1\\ \Leftrightarrow\left|3x\right|=2x+1\)

th1 :\(3x\ge0\Leftrightarrow x\ge0\Rightarrow\left|3x\right|=3x\)

\(\Rightarrow3x=2x+1\\ \Leftrightarrow3x-2x=1\\ \Leftrightarrow x=1\)

th2 : \(3x< 0\Leftrightarrow x< 0\Rightarrow\left|3x\right|=-3x\)

\(\Rightarrow-3x=2x+1\\ \Leftrightarrow-3x-2x=1\\ \Leftrightarrow-5x=1\\ \Leftrightarrow x=-\dfrac{1}{5}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(\sqrt{x^2}=7\\ \Leftrightarrow\left|x\right|=7\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)

 

Answer 2

`sqrt{9x^2} =2x+1`

`<=>sqrt{(3x)^2} = 2x+1`

`<=> |3x| = 2x+1`

\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\left(đk:x\ge0\right)\\-3x=2x+1\left(đk:x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2x=1\\-3x-2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{1}{5}\left(tm\right)\end{matrix}\right.\)

`sqrt{x^4} = 7`

`<=> x^2 = 7`

`<=> x = +- sqrt{7}`

`sqrt{x^2} = |-8|`

`<=> x = |-8|`

`=>x=8`

`sqrt{x^4} =9`

`=>x^2 = 9`

`=> x=+- 3`

`sqrt{x^2} = 7`

`=>|x| =7`

`=> x = +-7`

Answer 3

(1)<=> 3x=2x+1<=>x=1

(2) <=>x^2=7<=>x=\(\sqrt{7};x=-\sqrt{7}\)

(3)<=>x=7

(4)<=>x=8

(5)<=>x^2=9<=>x=3;x=-3

Answer 4

(1)<=> 3x=2x+1<=>x=1

(2) <=>x^2=7<=>x=$\sqrt{7};x=-\sqrt{7}$

(3)<=>x=7

(4)<=>x=8

(5)<=>x^2=9<=>x=3;x=-3