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RSB [31]
3 years ago
15

Va rog e urgent am nevoie de raspuns

Mathematics
1 answer:
Lady_Fox [76]3 years ago
3 0

Answer: 4

<u>Explanation:</u>

f(x) = 2x - 1

f(√2) = 2√2 - 1

f(1) = 2(1) - 1

     =  2 - 1

     =     1

f(√3)  = 2√3 - 1

*******************************************************

\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}

= \frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1}(\frac{\sqrt{2}+1}{\sqrt{2}+1})+\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}})

= \frac{4+2\sqrt{2}-2\sqrt{2}-2}{2 - 1} + \frac{6 +2\sqrt{6}-2\sqrt{6}-4}{3-2}

= \frac{2}{1} +\frac{2}{1}

= 4

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