Cho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2aCho 0 ≤ a,b,c ≤ 1. CMR 2 a 3 2a3 + 2 b 3 2b3 + 2 c 3 2c3 ≥ ≥ 3 + a 2 b + b 2 c + c 2 a a2b+b2c+c2a
Bất phương trình bậc nhất một ẩn.
\(P=\frac{4x+2}{2\left(x^2+2\right)}=\frac{-\left(x^2+2\right)+\left(x^2+4x+4\right)}{2\left(x^2+2\right)}=-\frac{1}{2}+\frac{\left(x+2\right)^2}{2\left(x^2+2\right)}\ge-\frac{1}{2}\)
\(P_{min}=-\frac{1}{2}\) khi \(x=-2\)
\(P=\frac{x^2+2-\left(x^2-2x+1\right)}{x^2+2}=1-\frac{\left(x-1\right)^2}{x^2+2}\le1\)
\(P_{max}=1\) khi \(x=1\)
Với mọi a;b;c ta luôn có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}.\left(\frac{3}{2}\right)^2=\frac{3}{4}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)
ĐKXĐ: \(x\ne4\)
Ta có: \(\frac{2x}{x-4}< 2\)
\(\Leftrightarrow2x< 2\left(x-4\right)\)
\(\Leftrightarrow2x< 2x-8\)
\(\Leftrightarrow2x-2x+8< 0\)
hay 8<0(vô lý)
Vậy: \(S=\varnothing\)
Ta có: \(5x-2\ge2x+4\)
\(\Leftrightarrow5x-2x\ge4+2\)
\(\Leftrightarrow3x\ge6\)
hay \(x\ge2\)
Vậy: S={x|\(x\ge2\)}
Mình nhìn lộn cái ≥ 2 nên mình không hiểu nên hỏi! Xin lỗi các bạn, không có ≥ 2 nha!
Không tồn tại min hay max của biểu thức S
Chỉ tồn tại min khi điều kiện x là \(0< x\le2\)
\(\frac{x^2-2x+9}{2\left(x-3\right)\left(x+1\right)}\le0\) \(\left(x\ne3;-1\right)\)
\(\Leftrightarrow\frac{\left(x-1\right)^2+8}{2\left(x-1\right)^2-8}\le0\)
Mà \(\left(x-1\right)^2+8>0\) \(\forall x\ne-1;3\)
\(\Rightarrow2\left(x-1\right)^2-8< 0\Leftrightarrow2\left(x-1\right)^2< 8\)
\(\Leftrightarrow\left(x-1\right)^2< 4\Leftrightarrow-1< x< 3\)
Vậy \(-1< x< 3\)
\((2x-1)(3-2x)(1-x)>0 \)
\(\Leftrightarrow\)\(4x^3-12x^2+11x-3>0\)
Bấm máy tính, ta được: \(\left[{}\begin{matrix}\frac{1}{2}< x< 1\\\frac{3}{2}< x\end{matrix}\right.\)
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