Question

Repede dau coroană,maine am mate si daca vin cu ea nefacuta primesc 4

Answer 1

 

Sub fiecare radical vom gasi sume sau diferente de 2 patrate

care apartin unor triplete de numere pitagorice.

De exemplu:

3² + 4² = 5²

6² + 8² = 10²

5² + 12² = 13²

8² + 15² = 17²

$\displaystyle\bf\\a)\\\\\sqrt{5^2+\frac{5^2\cdot3^2}{4^2}}=\\\\\\=\sqrt{\frac{5^2\cdot4^2+5^2\cdot3^2}{4^2}}=\\\\\\=\sqrt{\frac{5^2(4^2+3^2)}{4^2}}=\\\\\\=\sqrt{\frac{5^2\cdot5^2}{4^2}}=\\\\(deoarece~3^2+4^2=5^2,~~3;~4~si~5~fiind~numere~pitagorice)\\\\=\frac{5\cdot5}{4}=\boxed{\bf\frac{25}{4}}$

.

$\displaystyle\bf\\b)\\\\\sqrt{3^2+\frac{3^4}{2^4}}=\\\\\\=\sqrt{\frac{3^2\cdot2^4+3^4}{2^4}}=\\\\\\=\sqrt{\frac{3^2\cdot(2^2)^2+3^4}{2^4}}=\\\\\\=\sqrt{\frac{3^2\cdot4^2+3^4}{2^4}}=\\\\\\=\sqrt{\frac{3^2(4^2+3^2)}{2^4}}=\\\\\\=\sqrt{\frac{3^2\cdot5^2}{2^4}}=\frac{3\cdot5}{2^2}=\boxed{\bf\frac{15}{4}}$

.

$\displaystyle\bf\\c)\\\\\sqrt{7^2+\frac{7^2\cdot4^2}{3^2}}=\\\\\\=\sqrt{\frac{7^2\cdot3^2+7^2\cdot4^2}{3^2}}=\\\\\\=\sqrt{\frac{7^2(3^2+4^2)}{3^2}}=\\\\\\=\sqrt{\frac{7^2\cdot5^2}{3^2}}=\frac{7\cdot5}{3}=\boxed{\bf\frac{35}{3}}$

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$\displaystyle\bf\\d)\\\\\sqrt{4^2+\frac{4^2\cdot5^2}{12^2}}=\\\\\\=\sqrt{\frac{4^2\cdot12^2+4^2\cdot5^2}{12^2}}=\\\\\\=\sqrt{\frac{4^2(12^2+5^2)}{12^2}}=\\\\\\=\sqrt{\frac{4^2\cdot13^2}{12^2}}=\\\\(deoarece~12^2+5^2=13^2,~~5;~12~si~13~fiind~numere~pitagorice)\\\\=\frac{4\cdot13}{12}=\boxed{\bf\frac{13}{3}}$

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$\displaystyle\bf\\e)\\\\\sqrt{\frac{3^2\cdot13^2}{5^2}-3^2}=\\\\\\=\sqrt{\frac{3^2\cdot13^2-3^2\cdot5^2}{5^2}}=\\\\\\=\sqrt{\frac{3^2(13^2-5^2)}{5^2}}=\\\\\\=\sqrt{\frac{3^2\cdot12^2}{5^2}}=\frac{3\cdot12}{5}=\boxed{\bf\frac{36}{5}}$

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$\displaystyle\bf\\f)\\\sqrt{\frac{5^2\cdot8^2}{3^2}+5^2\cdot2^2}=\\\\=\sqrt{\frac{5^2\cdot8^2+5^2\cdot2^2\cdot3^2}{3^2}}=\\\\=\sqrt{\frac{5^2\cdot8^2+5^2\cdot(2\cdot3)^2}{3^2}}=\\\\=\sqrt{\frac{5^2\cdot8^2+5^2\cdot6^2}{3^2}}=\\\\=\sqrt{\frac{5^2(8^2+6^2)}{3^2}}=\\\\=\sqrt{\frac{5^2\cdot10^2}{3^2}}=\\\\(deoarece~8^2+6^2=10^2,~~6;~8~si~10~fiind~numere~pitagorice)\\\\=\frac{5\cdot10}{3}=\boxed{\bf\frac{50}{3}}$

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$\displaystyle\bf\\g)\\\sqrt{\frac{7^2\cdot8^2}{15^2}+7^2}=\\\\\\=\sqrt{\frac{7^2\cdot8^2+7^2\cdot15^2}{15^2}}=\\\\\\=\sqrt{\frac{7^2(8^2+15^2)}{15^2}}=\\\\\\=\sqrt{\frac{7^2\cdot17^2}{15^2}}=\\\\(deoarece~8^2+15^2=17^2,~~8;~15~si~17~fiind~numere~pitagorice)\\\\=\frac{7\cdot17}{15}=\boxed{\bf\frac{119}{15}}$

 

 

Answer 2

$\it a)\ \sqrt{5^2+\dfrac{5^2\cdot3^2}{4^2}}=\sqrt{5^2(1+\dfrac{9}{16})}=5\sqrt{\dfrac{25}{16}}=5\cdot\dfrac{5}{4}=\dfrac{25}{4}\\ \\ \\ b)\ \sqrt{3^2+\dfrac{3^4}{2^4}}=\sqrt{3^2(1+\dfrac{9}{16})}=3\sqrt{\dfrac{25}{16}}=3\cdot\dfrac{5}{4}=\dfrac{15}{4}\\ \\ \\ c)\ \sqrt{7^2+\dfrac{7^2\cdot4^2}{3^2}}=\sqrt{7^2(1+\dfrac{16}{9})}=7\sqrt{\dfrac{25}{9}}=7\cdot\dfrac{5}{3}=\dfrac{35}{3}$