Question

Dau coroana celui mai bun răspuns!

Answer 1

$4.\sqrt{x^2}=(-\sqrt{11})^2\\C.E.:x^2\geq 0\\\sqrt{x^2}=(-\sqrt{11})^2<=>|x|=11=>x_1=11\\x_2=-11\leq 0=>x_2-N.C.\\x\in\{11\}\\\\5.2x^2-mx+m=0\\x_1=x_2\\x_1,x_2\in R\\=>\Delta=0<=>(-m)^2-4*2*m=0<=>m^2-8m=0<=>m(m-8)=0=>m_1=0,m_2=8\\Pt.m\in\{0,8\}->x_1=x_2\in R$

$6.1+3+3^2+...+3^{99}\\3=\sqrt{1*3^2}<=>3=\sqrt{9}<=>3=3(A)=>$

=>1+3+3²+...+3⁹⁹ este o progresie geometrica cu primul termen b₁=1

b₂=b₁q<=>3=q=>ratia este q=3

Suma primilor n termeni ai unei progresii geometrice este data de formula:$S_n=\frac{b_1(q^n-1)}{q-1}$,unde n=100 si q=3

$S_{100}=\frac{b_1(q^{n}-1)}{q-1}=\frac{1(3^{100}-1)}{3-1} =\frac{3^{100}-1}{2} =>1+3+3^2+...+3^{99}=\frac{3^{100}-1}{2}$