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Edric hasn't added any friends yet.
21-04-2023 - 21:34
Ta có: $\frac{1}{a^2+b+c}=\frac{1}{a^2-a+3}=\frac{(a-3)(a-1)^2}{9(a^2-a+3)}+\frac{4-a}{9}\leqslant \frac{4-a}{9}$ Tương tự rồi cộng lại