Question

Varianta 1 va rog ajutati-ma❤ ofer 40 puncte si coroana❤

Answer 1

Răspuns:

Explicație pas cu pas:

a) (5t+10)²=(5t)²+2·5t·20z+(20z)²=25t²+200tz+400z²

2) 4002·3998=(4000+2)·(4000-2)=4000²-2²=16000000-4=15999996.

3) (t-1)³-(t+2)³=((t-1)-(t+2))((t-1)²+(t-1)(t+2)+(t+2)²)=(t-1-t-2)(t²-2t+1+t²+t-2+t²+4t+4)=-3·(3t²+3t+3)=-3·3·(t²+t+1)=-9·(t²+t+1).

4) (x+2)(x+3)-x-2=(x+2)(x+3)-(x+2)=(x+2)(x+3-1)=(x+2)(x+2)=(x+2)².

5) $\frac{3x^{2}-4}{x^{3}-8} :\frac{\sqrt{3} x+2}{x^{2}-4}=\frac{(\sqrt{3}x)^{2}-2^{2} }{x^{3}-2^{3}}: \frac{\sqrt{3} x+2}{x^{2}-2^{2}}=\frac{(\sqrt{3}x-2)(\sqrt{3}x+2)}{(x-2)(x^{2}+2x+2^{2})} *\frac{(x-2)(x+2)}{\sqrt{3}x+2} =\frac{(\sqrt{3}x-2)(x+2)}{x^{2}+2x+4}$

6) x²-8x+20=0, ⇒ ecuatia aceasta nu are solutii, cred e gresita

daca era x²-8x-20=0,⇒x²-2·x·4+4²-4²-20=0, ⇒(x-4)²-36=0, ⇒(x-4)²-6²=0, ⇒(x-4-6)(x-4+6)=0, ⇒(x-10)(x+2)=0, de unde rezulta x-10=0 sau x+2=0

De unde x=10 sau x=-2. Raspuns: S={-2; 10}

Daca rezolvam ecuatia x²-8x+20=0, ⇒

x²-2·x·4+4²+4=0, ⇒(x-4)²+4=0 Ecuatia nu are solutii, deoarece expresia (x-4)²+4≥4., adica nu poate fi 0.