Question

\(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}=0\)

Answer 1

Ta có : \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^{50}\ge0\forall x\\\left(y+\frac{1}{3}\right)^{40}\ge0\forall y\end{cases}}\Rightarrow\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}\ge0\forall x;y\)

Khi đó \(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}=0\)

<=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y+\frac{1}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)

Vậy x = 1/2 ; y = -1/3

Answer 2

Ta có: \(\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^{50}=0\\\left(y+\frac{1}{3}\right)^{40}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)

Answer 3

Vì \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^{50}\\\left(y+\frac{1}{3}\right)^{40}\end{cases}}\ge0\forall x,y\Rightarrow\left(x-\frac{1}{2}\right)^{50}+\left(y+\frac{1}{3}\right)^{40}\ge0\)

Dấu "=" xảy ra <=> x = 1/2 ; y = -1/3

Vậy x = 1/2 ; y = -1/3

Answer 4

cảm ơn vì tất cả

Answer 5

27^2×8^5/6^6×32^2+3^4×4^4/2^2×6^2