Question

1.Calculati $\frac{4}{2+ \sqrt{2} } -5+2 \sqrt{2}$
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2.Fie F:R⇒R,f(x)=1+3x.Determinati valorile reale ale lui x,care sunt mai mici decit valorile corespunzatoare ale functiei f.
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3.Scrieti expresia $\frac{x-3}{x+3} - \frac{x+3}{3-x} + \frac{12x}{ x^{2}-9 }$ sub forma de fractie algebrica ireductibila pe domenilor valorilor admisibile
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4.Fie functia F:R⇒R,f(x)=-x²+2mx-(m-2)².Determinati valorile reale ale lui m,p/u care functia f are cel putin un zerou.


Va rog sa ma ajutati mult pentru ca am examen.

Answer 1

1. Trebuie sa amplifici fractia cu 2-√2 (conjugata lui 2+√2)
$\frac{4(2- \sqrt{2} )}{(2+ \sqrt{2} )(2- \sqrt{2} )}= \frac{4(2- \sqrt{2} )}{4-2}= \frac{4(2- \sqrt{2} )}{2}=2(2- \sqrt{2} )=4-2 \sqrt{2}$
4-2√2-5+2√2=4-5= -1
2. x<f(x)
x<1+3x ⇔ -2x<1 ⇔ x>-1/2 ⇔x∈(-1/2,+∞)
3.$\frac{x-3}{x+3}- \frac{x+3}{3-x}+ \frac{12x}{ x^{2}-9 }= \frac{x-3}{x+3}+ \frac{x+3}{x-3}+ \frac{12x}{(x-3)(x+3)}= \frac{(x-3)^2+(x+3)^2+12x}{(x-3)(x+3)}=$$\frac{ x^{2}-6x+9+ x^{2}+6x+9+12x}{(x-3)(x+3)}= \frac{2 x^{2}+12x+18 }{(x-3)(x+3)}= \frac{2( x^{2} +6x+9)}{(x-3)(x+3)}= \frac{2(x+3)^2}{(x-3)(x+3)}= \frac{2(x+3)}{x-3}$