Daily math problems for May 28, 2026 — Algebraic identities, Calculus, Logaritmi

$(x+y{)}^2=(x^2+y^2)+2xy=13+12=25$, so $\mathrm{\mid }x+y\mathrm{\mid }=5$.

$\bullet$ Simplify the limit by factoring numerator and denominator:

$$\underset{k\to x}{\lim }\frac{(k-4)(k+4)}{(k-4)(k-1)}=\frac{x+4}{x-1}$$

$\bullet$ The determinant has a row of zeros (with ${\log }_{3}1=0$), so it contributes $0$:

$$f(x)=\frac{x+4}{x-1}$$

$\bullet$ Evaluate at $x=4$ and $x=0$:

$$f(4)=\frac{8}{3},f(0)=-4$$

$\bullet$ Sum them:

$$f(4)+f(0)=\frac{8}{3}-4=-\frac{4}{3}$$

Folosind schimbarea de bază, ${\log }_k(k+1)={\displaystyle \frac{\ln (k+1)}{\ln k}}$. Produsul telescopează: ${\displaystyle \frac{\ln 3}{\ln 2}}\cdot {\displaystyle \frac{\ln 4}{\ln 3}}\cdots {\displaystyle \frac{\ln 256}{\ln 255}}={\displaystyle \frac{\ln 256}{\ln 2}}={\log }_{2}256=8$.