giải pt: a) căn(2x²-1)+căn(x²-3x-2)=căn(2x²+2x+3)+căn(x²-x+2)
b) [căn(3x+1)-căn(x+2)] . [căn(3x²+7x+2)+4]=4x-2
c) căn(x-1) = (x²+1)/2
d) căn(x-1)+căn(x-2)=x
giải pt: a) căn(2x²-1)+căn(x²-3x-2)=căn(2x²+2x+3)+căn(x²-x+2) b) [căn(3x+1)-căn(x+2)] . [căn(3x²+7x+2)+4]=4x-2 c) căn(x-1) =
By Hadley
`a) \sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}` (1)
ĐKXĐ: `x>= (3+\sqrt{17})/2` hoặc `x<=\sqrt{2}/2`
`(1)<=> (\sqrt{2x^2+2x+3}-\sqrt{2x^2-1})+(\sqrt{x^2-x+2}-\sqrt{x^2-3x-2})=0`
`<=> (2x^2+2x+3-2x^2+1)/(\sqrt{2x^2+2x+3}+\sqrt{2x^2-1})+(x^2-x+2-x^2+3x+2)/(\sqrt{x^2-x+2}+\sqrt{x^2-3x-2})=0`
`<=> (2x+4)/(\sqrt{2x^2+2x+3}+\sqrt{2x^2-1})+(2x+4)/(\sqrt{x^2+x+2}+\sqrt{x^2-3x-2})=0`
`<=> (2x+4)(1/(\sqrt{2x^2+2x+3}+\sqrt{2x^2-1})+1/(\sqrt{x^2+x+2}+\sqrt{x^2-3x-2}))=0`
Do `1/(\sqrt{2x^2+2x+3}+\sqrt{2x^2-1})+1/(\sqrt{x^2+x+2}+\sqrt{x^2-3x-2})>0` với `AAx`
`-> 2x+4=0`
`<=> x=-2` ™
Vậy `S={-2}`
`b) (\sqrt{3x+1}-\sqrt{x+2})(\sqrt{3x^2+7x+2}+4)=4x-2` (2)
ĐKXĐ: `x>=-1/3`
Đặt `\sqrt{3x+1}=a; \sqrt{x+2}=b` (`a,b>=0`)
`-> \sqrt{3x^2+7x+2}=ab`
`4x-2=2(a^2-b^2)`
`(2)-> (a-b)(ab+4)=2(a^2-b^2)`
`<=> (a-b)(ab+4)=2(a-b)(a+b)`
`<=> (a-b)(ab+4-2a-2b)=0`
`<=> (a-b)(a-2)(b-2)=0`
`<=>`\(\left[ \begin{array}{l}a=b\\a=2\\b=2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}3x+1=x+2\\3x+1=4\\x+2=4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\\x=2\end{array} \right.\) ™
Vậy `S={1/2;1;2}`
`c) \sqrt{x-1}=(x^2+1)/2` (3)
ĐK: `x>=1`
`(3)<=> 2\sqrt{x-1}=x^2+1`
Có: `2\sqrt{x-1}<=x-1+1=x`
`-> x^2+1<=x`
`<=> x^2-x+1<=0`
`<=> (x-1/2)^2+3/4<=0` (vô lý)
Vậy `S=∅`
`d) \sqrt{x-1}+\sqrt{x-2}=x` (4)
`ĐK: `x>=1`
Đặt `\sqrt{x-1}=a; \sqrt{x-2}=b` (`a,b>=0`)
`-> a^2=x-1; b^2=x-2`
`-> x=(a^2+b^2+3)/2`
`(4)-> a+b=(a^2+b^2+3)/2`
`<=> 2a+2b=a^2+b^2+3`
`<=> (a^2-2a+1)+(b^2-2b+1)+1=0`
`<=> (a-1)^2+(b-1)^2+1=0` (vô lý)
Vậy `S=∅`
Đáp án:
Giải thích các bước giải:
a) ĐKXĐ $: 2x² – 1 ≥ 0 ⇔ x ≤ – \dfrac{\sqrt{2}}{2} ; x ≥ \dfrac{\sqrt{2}}{2}$
$ x² – 3x – 2 ≥ 0 ⇔ x ≤ \dfrac{3 – \sqrt{17}}{2} ; x ≥ \dfrac{3 + \sqrt{17}}{2}$
Kết hợp lại $ x ≤ – \dfrac{\sqrt{2}}{2}; x ≥ \dfrac{3 + \sqrt{17}}{2}$
$PT ⇔ (\sqrt{2x² – 1} – \sqrt{2x² + 2x + 3}) + (\sqrt{x² – 3x – 2} – \sqrt{x² – x + 2}) = 0$
$ ⇔ \dfrac{(2x² – 1) – (2x² + 2x + 3)}{\sqrt{2x² – 1} + \sqrt{2x² + 2x + 3}} + \dfrac{(x² – 3x – 2) – (x² – x + 2)}{\sqrt{x² – 3x – 2} + \sqrt{x² – x + 2}}) = 0$
$ ⇔ – \dfrac{2(x + 2)}{\sqrt{2x² – 1} + \sqrt{2x² + 2x + 3}} – \dfrac{2(x + 2)}{\sqrt{x² – 3x – 2} + \sqrt{x² – x + 2}}) = 0$
$ ⇔ – 2(x + 2)(\dfrac{1}{\sqrt{2x² – 1} + \sqrt{2x² + 2x + 3}} + \dfrac{1}{\sqrt{x² – 3x – 2} + \sqrt{x² – x + 2}}) = 0$
$ ⇔ x + 2 = 0 ⇔ x = – 2 (TM)$
KL : PT có nghiệm duy nhất $x = – 2$
b) ĐKXĐ $: 3x + 1 ≥ 0 ⇔ x ≥ – \dfrac{1}{3}; x + 2 ≥ 0 ⇔ x ≥ – 2$
$ 3x² + 7x + 2 = (3x + 1)(x + 2) ≥ 0 ⇔ x ≤ – 2; x ≥ – \dfrac{1}{3}$
Kết hợp lại $: x ≥ – \dfrac{1}{3}$
$ PT ⇔ [(3x + 1) – (x + 2)](\sqrt{3x² + 7x + 2} + 4) = 2(2x – 1)(\sqrt{3x + 1} + \sqrt{x + 2})$
$ ⇔ (2x – 1)(\sqrt{3x² + 7x + 2} + 4) – 2(2x – 1)(\sqrt{3x + 1} + \sqrt{x + 2}) = 0$
$ ⇔ (2x – 1)[\sqrt{(3x+ 1)(x + 2)} + 4 – 2\sqrt{3x + 1} – 2\sqrt{x + 2}] = 0$
$ ⇔ (2x – 1)(\sqrt{3x + 1} – 2)(\sqrt{x + 2} – 2) = 0$
– TH1 $: 2x – 1 = 0 ⇔ x = \dfrac{1}{2} (TM)$
– TH2 $: \sqrt{3x + 1} – 2 = 0 ⇔ 3x + 1 = 4 ⇔ x = 1 (TM)$
– TH3 $: \sqrt{x + 2} – 2 = 0 ⇔ x + 2 = 4 ⇔ x = 2 (TM)$
KL : PT có 3 nghiệm $x = \dfrac{1}{2}; x = 1; x = 2$
c) ĐKXĐ $: x – 1 ≥ 0 ⇔ x ≥ 1$
$PT ⇔ 2\sqrt{x – 1} = x² + 1 ≥ 2x $
$ ⇔ \sqrt{x – 1} ≥ x ⇔ x – 1 ≥ x² ⇔ x² – x + 1 ≤ 0$
$ ⇔ (x – \dfrac{1}{2})² + \dfrac{3}{4} ≤ 0$ ( vô lý)
$ ⇒ PT $ vô nghiệm
d) ĐKXĐ $: x – 1 ≥ 0; x – 2 ≥ 0 ⇒ x ≥ 2$
$ PT ⇔ 2\sqrt{x – 1} + 2\sqrt{x – 2} = 2x$
$ ⇔ (x – 1) – 2\sqrt{x – 1} + 1 + (x – 2) – 2\sqrt{x – 1} + 1 + 1 = 0$
$ ⇔ (\sqrt{x – 1} – 1)² + (\sqrt{x – 12} – 2)² + 1 = 0$ (vô lý)
$ ⇒ PT $ vô nghiệm