Question

Cine imi rezolva si mie pe 3,4,5 dau coroana si 50 de puncte. Mersi!

Answer 1

Răspuns:

3.

I: 4x/9; II: (60/100)·(5x/9); III4

$x=\frac{4x}{9} +\frac{60}{100} *\frac{5x}{9} +4$

$x=\frac{4x}{9} +\frac{x}{3} +4$   /·9

9x=4x+3x+36

2x=36  ⇒ x=18

4. a=√2; b=√3

a) $b^{2a^{2} } >a^{2b^{2} }$

$\sqrt{3}^{2\sqrt{2}^{2} } >\sqrt{2}^{2\sqrt{3}^{2} }$

$\sqrt{3}^{2*2} >\sqrt{2}^{2*3}}$

$3^{2} >2^{3}$ ⇒ 9>8 Adevarat ⇒ $b^{2a^{2} } >a^{2b^{2} }$

b)$\frac{1}{\sqrt{a^{2}-b}}+\frac{1}{\sqrt{a^{2}+b}}=\sqrt{6}$

$\frac{\sqrt{a^{2}+b})1}{\sqrt{a^{2}-b}}+\frac{\sqrt{a^{2}-b})1}{\sqrt{a^{2}+b}}=\sqrt{6}$

$\frac{\sqrt{a^{2}+b}+{\sqrt{a^{2}-b}}}{\sqrt{a^{2}+b}*{\sqrt{a^{2}-b}}}=\sqrt{6}$

$\frac{\sqrt{a^{2}+b}+{\sqrt{a^{2}-b}}}{\sqrt{(a^{2}+b)(a^{2}-b)}}=\sqrt{6}$

$\frac{\sqrt{a^{2}+b}+{\sqrt{a^{2}-b}}}{\sqrt{(a^{4}-b^{2})}}=\sqrt{6}$

$\frac{\sqrt{2+\sqrt{3} }+{\sqrt{2-\sqrt{3}}}}{\sqrt{(4-3})}}=\sqrt{6}$

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

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

$\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{6}$   /·$\sqrt{2}$

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

$2\sqrt{3}= 2\sqrt{3}$ Adevarat ⇒ $\frac{1}{\sqrt{a^{2}-b}}+\frac{1}{\sqrt{a^{2}+b}}=\sqrt{6}$

5. E(x)=x³+x²-x+2

a, b, c=? a.i. E(x) =(x+2)(ax²+bx+c)

(x+2)(ax²+bx+c)=ax³+bx²+cx+2ax²+2bx+2x=ax³+x²(2a+b)+x(2b+c)+2c ⇒

a=1

2a+b=1 ⇒ b=-2a ⇒ b=-1

2b+c=-1

2c=2 ⇒ c=1

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