giải pt √(x²- 9) + √(x²- 6x+ 9)= 0 Question giải pt √(x²- 9) + √(x²- 6x+ 9)= 0 in progress 0 Môn Toán Adela 4 years 2020-10-13T21:06:36+00:00 2020-10-13T21:06:36+00:00 2 Answers 118 views 0
Answers ( )
$\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0_{}$ $ĐK:_{}$ $ĐK:_{}$ $-3_{}$ $\geq$ $x_{}$ $\geq3$
$⇔\sqrt{x^2-9}=-\sqrt{x^2-6x+9}_{}$
$⇔x^2-9=x^2-6x+9_{}$
$⇔6x=18_{}$
$⇔x=3_{}$ $(TMĐK)_{}$
$Vậy_{}$ $x=3_{}$
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`
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