Cho các số thực dương $a,b,c$ thỏa mãn $(a+b)^{2}+2c^{2}\geq 2$
Tìm min $P=5(a^{2}+b^{2}+c^{2})-(a+b+\sqrt{2}c)^{2}-\sqrt{\frac{(a+b)^{2}}{2}+c^{2}}$
uk.vào đề thi thử ĐH mà dễ nhỉ:)) –  Bloody's Rose 25-06-16 09:06 PM
bài này vừa vào đề thi thử DH của tỉnh em nek chị :D –  ๖ۣۜJinღ๖ۣۜKaido 25-06-16 05:45 PM
$\mathit{P \ge 5\frac{(a+b)^2+2c^2}{2}-2\bigg[(a+b)^2+2c^2 \bigg]-\sqrt{\frac{(a+b)^2+2c^2}{2}} \\=\frac{(a+b)^2+2c^2}{2}-\sqrt{\frac{(a+b)^2+2c^2}{2}} \\ Dat \; \sqrt{\frac{(a+b)^2+2c^2}2}=t,t \ge 1 \\ \Rightarrow P \ge f(t)=t^2-t \\=t(t-1) \ge 0 \\ \mathbf{Mặt\;khác\;khi\;thay\;}a=b=\frac 12,c=\sqrt{\frac 12}}\; \mathbf{thì\;} P=0 \\ \mathbf{Vậy} \min P=0$
vái sư phụ tay j mà tay chứ a cài kiểu j z chỉ e vs –  ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 26-06-16 03:41 PM
chữ viết tay đó e @@ –  tran85295 26-06-16 11:00 AM
cái số 2 n' kì kì –  ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 26-06-16 07:26 AM
chuẩn r:)) –  Bloody's Rose 25-06-16 09:05 PM

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