giải pt √(x²- 9) + √(x²- 6x+ 9)= 0

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giải pt
√(x²- 9) + √(x²- 6x+ 9)= 0

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Adela 1 year 2020-10-13T21:06:36+00:00 2 Answers 102 views 0

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    0
    2020-10-13T21:07:38+00:00

    $\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0_{}$ $ĐK:_{}$ $ĐK:_{}$ $-3_{}$ $\geq$ $x_{}$ $\geq3$ 

    $⇔\sqrt{x^2-9}=-\sqrt{x^2-6x+9}_{}$

    $⇔x^2-9=x^2-6x+9_{}$

    $⇔6x=18_{}$

    $⇔x=3_{}$ $(TMĐK)_{}$

    $Vậy_{}$ $x=3_{}$

    0
    2020-10-13T21:08:06+00:00

    Đáp án:

    Ta có : 

    ` \sqrt{x^2 – 9} + \sqrt{x^2 – 6x + 9} = 0`          `(ĐKXĐ : x ≥ 3` hoặc `x ≤ -3 )`

    `<=> \sqrt{x^2 – 9} = -\sqrt{x^2 – 6x + 9}`

    `<=> (\sqrt{x^2 – 9})^2 = (-\sqrt{x^2 – 6x + 9})^2`

    `<=> x^2 – 9 = x^2 – 6x + 9`

    `<=> x^2 – 9 – x^2 + 6x – 9 = 0`

    `<=> 6x – 18 = 0`

    `<=> 6x = 18`

    `<=> x = 3`

    Giải thích các bước giải:

     

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