Phuong Math
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\(\displaystyle \frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}=\frac{2}{1+\sqrt{x}}\)
Last edited:
\(\displaystyle \frac{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}=\frac{2}{1+\sqrt{x}}\)
I edited. Can you help me, plz.?Even with [ tex ] and [ \tex ] tags this doesn't make any sense. You are missing the denominator to one of your fractions.
Are you have any good idea for that problem??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.
http://www.codecogs.com/latex/eqneditor.phpPlease give me link to Latex.
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?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
So sqrt(0) >0? How much bigger?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.