Question

Ajutor va roog , macar primul subiect !

Answer 1

................................................................

Answer 2

Subiectul 1:
1. a) $8^{3}+ 16^{2}-24x+40^{5}=8x*( x^{2}+2x-3+5x^{5})$
b) $15xy^{2}+45 x^{2}y=15xy*(y+3x)$
c) $\frac{1}{3} x^{2}+ \frac{1}{3}x+ \frac{1}{3}= \frac{1}{3}*( x^{2} +x+1)$
d) 3x*(x+2)-6(x+2)=(3x-6)*(x+2)=3(x-2)*(x+2)

2. (2x-1)(2x+1)+(x+2)(x-2)=$4 x^{2}-1+ x^{2}-4= 5x^{2}-5$
b) $(x+3 \sqrt{2})^{2}+(x-3 \sqrt{2} )^{2} = x^{2} +6 \sqrt{2}x+18+ x^{2} -6 \sqrt{2}x+18$$=2x^{2} +36$
c) (x-3)²+(x-3)(x+3)=(x-3)*(x-3+x+3)=(x-3)*2x=2x*(x-3)
d) (x+y-3)²=x²+y²=(-3)²+2xy+2x*(-3)+2y*(-3)=x²+y²+9+2xy-6x-6y

3. a) $x^{2} - \frac{81}{64}=(x- \frac{9}{8})*(x+ \frac{9}{8}$
b) $(x+ \sqrt{2}) ^{2} -32=(x+ \sqrt{2}- \sqrt{32} )*(x+ \sqrt{2}+ \sqrt{32})=(x+ \sqrt{2}-4 \sqrt{2})$ $*(x+ \sqrt{2}+4 \sqrt{2}=(x-3 \sqrt{2})*(x+5 \sqrt{2} )$
c) $x-2 \sqrt{3}x+3-(y+ \sqrt{3} ^{2}=x-2 \sqrt{3}+3-(y^{2} +2 \sqrt{3}y+3)$$=x-2 \sqrt{3}x+3-y^{2}-2 \sqrt{3}y-3=x-2 \sqrt{3}x-y^{2} -2 \sqrt{3}y$
d) x²-11x+10=x²-x-10x+10=x*(x-1)-10(x-1)=(x-10)*(x-1)

Subiectul 2:
1. foto.
2. e(x)=x²+x+3
e(x)-x²-x-3=0
(e-1)x-x²-3=0
-x²+(e-1)x-3=0
x²-(e-1)x+3=0
$x= \frac{-(-(e-1))+ \sqrt{(-(e-1))^{2}-4*1*3 } }{2*1}$
$x= \frac{e-1+ \sqrt{(e-1)^{2}-12 } }{2}$
$x= \frac{e-1+ \sqrt{e^{2}-2e+1-12 } }{2}$
$x= \frac{e-1+ \sqrt{e^{2}-2e-11 } }{2}$
x∉ER

3. $x+ \frac{1}{x} \geq 2$
$x+ \frac{1}{x}-2 \geq 0$
$\frac{ x^{2}+12 }{x} \geq 0$
$\frac{ x^{2} -2x+1}{x} \geq 0$
$\frac{(x-1)^{2} }{x} \geq 0$
$\left \{ {{(x-1)^{2} \geq 0 } \atop {x\ \textgreater \ 0}} \right.$
$\left \{ {{(x-1)^{2} \leq 0 } \atop {x\ \textless \ 0}} \right.$
$\left \{ {{x∈ER} \atop {x\ \textgreater \ 0}} \right.$
$\left \{ {{x=1} \atop {x\ \textless \ 0}} \right.$
x∈(0,+∞)
x∈∅
x∈(0,+∞)