Solve equation

Phuong Math

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\(\displaystyle \frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}=\frac{2}{1+\sqrt{x}}\)
 
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[QUOTE = bestellen; 368408] x> 0
sqrt (x + 3) (x-1) (x + 1/3) + sqrt (3x + 1) (x-1) (x + 3) - 2 (1- sqrt (x)) (x + 3) (x + 1/3) = 0
sqrt (x + 3) (x-1) (x + 1/3)
(1 + sqrt (x)) + sqrt (3x + 1) (x-1) (x + 3) (1 + sqrt (x)) + 2 (x-1) (x + 3) (x + 1/3) = 0 => x = 1



sqrt (x + 3) (x + 1/3) (1 + sqrt (x)) + sqrt (3x + 1) (x + 3) (1 + sqrt (x)) + 2 (x + 3) (x + 1/3 ) = 0 không có rễ nên khi x> 0
trả lời: x = 1

[/ QUOTE]
Can you write with Latex? I don't understand your post.
 
Let a = SQRT(x + 3), b = SQRT(3x + 1), c = 1 + SQRT(x)
1/a + 1/b = 2/c ; leads to:
c(a + b) / (ab) = 2
Substitute back in: too UGLY to even attempt!

Anyhow, easy to see (by inspection) that x = 1.
Are you have any good idea for that problem??
 
x>0

\(\displaystyle \displaystyle{ \sqrt{x\, +\, 3\,}\, (x\, -\, 1)\,\left(x\, +\, \frac{1}{3}\right)\, +\, \sqrt{3x\, +\, 1\,}\,(x\, -\, 1)\, (x\, +\, 3)\, -\, 2\,\left(1\, -\, \sqrt{x\,}\right)\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right)\, =\, 0 }\)

\(\displaystyle \displaystyle{ \sqrt{x\, +\, 3\,}\, (x\, -\, 1)\,\left(x\, +\, \frac{1}{3}\right)\,\left(1\, +\, \sqrt{x\,}\right)\, +\, \sqrt{3x\, +\, 1\,}\,(x\, -\, 1)\, (x\, +\, 3)\, \left(1\, +\, \sqrt{x\,}\right)\, + \, 2\,(x\, -\, 1)\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right)\, =\, 0 }\)

\(\displaystyle \displaystyle{ (x\, -\, 1)\, \sqrt{x\, +\, 3\,}\, \left[ \left(x\, +\, \dfrac{1}{3}\right)\, \left(1\, +\, \sqrt{x\,}\right)\, +\, \sqrt{3x\, +\, 1\,}\,(x\, +\, 3)\, \left(1\, +\, \sqrt{x\,}\right)\, 2\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right) \right]\, =\, 0 }\)

\(\displaystyle \mbox{Then:}\)

\(\displaystyle \displaystyle{ x\, =\, 1\, \sqrt{x\, +\, 3\,}\, \left(x\, +\, \frac{1}{3}\right)\, \left(1\, +\, \sqrt{x\,}\right)\, +\, \sqrt{3x\, +\, 1\,}\, (x\, +\, 3)\, \left(1\, +\, \sqrt{x\,}\right)\, +\, 2\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right)\, =\, 0 }\)

\(\displaystyle \mbox{has no roots so when }\, x\, >\, 0\)

\(\displaystyle \mbox{Answer: }\, x\, =\, 1\)



View attachment 4803
has no roots so when x>0
Answer: x=1
Oops. I see it more time and detection your solution was wrong in the first line.
 
x>0

\(\displaystyle \displaystyle{ \sqrt{x\, +\, 3\,}\, (x\, -\, 1)\,\left(x\, +\, \frac{1}{3}\right)\, +\, \sqrt{3x\, +\, 1\,}\,(x\, -\, 1)\, (x\, +\, 3)\, -\, 2\,\left(1\, -\, \sqrt{x\,}\right)\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right)\, =\, 0 }\)

\(\displaystyle \displaystyle{ \sqrt{x\, +\, 3\,}\, (x\, -\, 1)\,\left(x\, +\, \frac{1}{3}\right)\,\left(1\, +\, \sqrt{x\,}\right)\, +\, \sqrt{3x\, +\, 1\,}\,(x\, -\, 1)\, (x\, +\, 3)\, \left(1\, +\, \sqrt{x\,}\right)\, + \, 2\,(x\, -\, 1)\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right)\, =\, 0 }\)

\(\displaystyle \displaystyle{ (x\, -\, 1)\, \sqrt{x\, +\, 3\,}\, \left[ \left(x\, +\, \dfrac{1}{3}\right)\, \left(1\, +\, \sqrt{x\,}\right)\, +\, \sqrt{3x\, +\, 1\,}\,(x\, +\, 3)\, \left(1\, +\, \sqrt{x\,}\right)\, 2\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right) \right]\, =\, 0 }\)

\(\displaystyle \mbox{Then:}\)

\(\displaystyle \displaystyle{ x\, =\, 1\, \sqrt{x\, +\, 3\,}\, \left(x\, +\, \frac{1}{3}\right)\, \left(1\, +\, \sqrt{x\,}\right)\, +\, \sqrt{3x\, +\, 1\,}\, (x\, +\, 3)\, \left(1\, +\, \sqrt{x\,}\right)\, +\, 2\, (x\, +\, 3)\, \left(x\, +\, \frac{1}{3}\right)\, =\, 0 }\)

\(\displaystyle \mbox{has no roots so when }\, x\, >\, 0\)

\(\displaystyle \mbox{Answer: }\, x\, =\, 1\)



View attachment 4803
has no roots so when x>0
Answer: x=1
The first, third and four line are not correct. But everyone makes mistakes. What baffles me is that YOUR last equations says that x=0. Then you say that there are NO roots when x>0. Then you say that x=1 is the answer. This contradicts that there are no roots when x>0. Are you suggesting that 1 is not greater than 0?
 
If we have equation
f1(x)*f2(X) = 0 then f1(x)=0 or f2(x)=0
in our example
f1(x) = x-1 => x=1
f2(x) = sqrt(x+3)........ > 0 (all sums > 0, x>0) never equal 0 => f2(x) no roots.
So sqrt(0) >0? How much bigger?
Note that if x=-3, then x+3=0
 
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