1,3x^ 4 -6x ^2 = 0 2,√x(√x+2)=+3

Đáp án:

Giải thích các bước giải:

1/

`3x^4 -6x^2 =0`

`<=> 3x^2(x^2 - 2) =0` 

`<=>` \(\left[ \begin{array}{l}x^2 =0\\x^2 -2 =0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x^2 = 2 \end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x=\pm \ \sqrt{2}\end{array} \right.\) 
Vậy `S ={0 ; \pm \sqrt{2} }`

2/

`\sqrt{x}(\sqrt{x}+2) =3`

`<=> (\sqrt{x})^2 + 2\sqrt{x}   =3`

`<=> (\sqrt{x})^2 + 2. \sqrt{x} .1 +1 = 4`
`<=> (\sqrt{x} +1)^2 =4`

`<=>`  \(\left[ \begin{array}{l}\sqrt{x} +1 =2\\\sqrt{x}+1 =-2\end{array} \right.\) 

`<=>`  \(\left[ \begin{array}{l}\sqrt{x} =1\\ \sqrt{x} =-3 \ ( L ) \end{array} \right.\) 

`<=> x =1`
Vậy `S ={1}` 

`1)3x^ 4 -6x ^2 = 0  `

`<=>3x^2(x^2-2)=0`

$⇔\left[\begin{matrix} 3x^2=0\\ x^2-2=0\end{matrix}\right.$

$⇔\left[\begin{matrix} x^2=0\\ x^2=2\end{matrix}\right.$

$⇔\left[\begin{matrix} x=0\\ x=±\sqrt2\end{matrix}\right.$

Vậy `x in {0;sqrt2;-sqrt2}`

`2)sqrtx(sqrtx+2)=3(x>=0)`

`<=>x+2sqrtx-3=0`

`<=>x+3sqrtx-sqrtx-3=0`

`<=>(x+3sqrtx)-(sqrtx+3)=0`

`<=>sqrtx(sqrtx+3)-(sqrtx+3)=0`

`<=>(sqrtx+3)(sqrtx-1)=0`

`<=>sqrtx-1=0(` vì ` sqrtx+3>0)`

`<=>sqrtx=1`

`<=>x=1,tm`
Vậy `x=1`