Ta thấy:\(\hept{\begin{cases}\left(2x-1\right)^2\ge0\\\left(3y+5\right)^2\ge0\end{cases}}\Rightarrow\hept{\begin{cases}3\left(2x-1\right)^2\ge0\\7\left(3y+5\right)^2\ge0\end{cases}}\)(với mọi x,y)

mà \(3\left(2x-1\right)^2+7\left(3y+5\right)^2=0\Rightarrow\hept{\begin{cases}3\left(2x-1\right)^2=0\\7\left(3y+5\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-1=0\\3y+5=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-5}{3}\end{cases}}}\)

Vậy....

NHỚ K ĐẤY

ta có 

\(\hept{\begin{cases}3\left(2x-1\right)^2\ge0\text{ }\forall x\\7\left(3y+5\right)^2\ge0\text{ }\forall y\end{cases}}\Rightarrow3\left(2x-1\right)^2+7\left(3y+5\right)^2\ge0\)

dấu bằng xảy ra khi \(\hept{\begin{cases}2x-1=0\\3y+5=0\end{cases}\Rightarrow x=\frac{1}{2}}\)

Vậy x=1/2

\(\left(2a^2-b\right)^2=\left(2a^2\right)^2-2.2a^2.b+b^2\)

Ta có x=14 suy ra x+1=15

Do đó thay x+1 vào H(x), ta được:

H(14)= x¹⁰ - (x+1)x⁹ +(x+1)x⁸-(x+1)x⁷+...+ (x+1)x² - (x+1)x + x+1

H(14)= x^10 - x^10 -x^9 +x^8- x^8-x^7 +....+ x^3 +x^2 -x^2-x+x +1

Hay H(14)=1

Đặt Q(x) = x⁹ - x⁸ + x⁷ - ... + x - 1 thì (x + 1) * Q(x) = (x + 1) * (x⁹ - x⁸ + x⁷ - ... + x - 1) = x¹⁰ - 1 \(\Rightarrow\left(14+1\right)\cdot Q\left(14\right)=14^{10}-1\)

Dễ thấy: H(x) = x¹⁰ - 15* Q(x) \(\Rightarrow H\left(14\right)=14^{10}-\left(14^{10}-1\right)=1\)

Xin cậu !

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)< \left(27-25\right)\left(27+25\right)=27^2-25^2=B\)

Vậy A < B.