Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. BĐT tương đương với \(6\left(a^2+b^2\right)-2ab+8-4\left(a\sqrt{b^2+1}+b\sqrt{a^2+1}\right)\ge0\)
\(\Leftrightarrow\left[a^2-4a\sqrt{b^2+1}+4\left(b^2+1\right)\right]+\left[b^2-4b\sqrt{a^2+1}+4\left(a^2+1\right)\right]\)\(+\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a-2\sqrt{b^2+1}\right)^2+\left(b-2\sqrt{a^2+1}\right)^2+\left(a-b\right)^2\ge0\)(đúng)
=> Đẳng thức không xảy ra
2. \(a^4+b^4+c^2+1\ge2a\left(ab^2-a+c+1\right)\)
\(\Leftrightarrow a^4+b^4+c^2+1\ge2a^2b^2-2a^2+2ac+2a\)
\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+\left(c^2-2ac+a^2\right)+\left(a^2-2a+1\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c-a\right)^2+\left(a-1\right)^2\ge0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)\(2a^4+1\ge2a^3+a^2\)
\(\Leftrightarrow2a^4-2a^3-a^2+1\ge0\)
\(\Leftrightarrow\left(a^4-2a^3+a^2\right)+\left(a^4-2a^2+1\right)\ge0\)
\(\Leftrightarrow\left(a^2-a\right)^2+\left(a^2-1\right)^2\ge0\)(luôn đúng)
"="<=>a=1
Ta có:\(2A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\)
\(2A=1-\dfrac{1}{11}=\dfrac{10}{11}\)
\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{9\cdot11}\right)\)
\(B=\dfrac{4}{1\cdot3}\cdot\dfrac{9}{2\cdot4}\cdot...\cdot\dfrac{100}{9\cdot11}\)
\(B=\dfrac{2\cdot2\cdot3\cdot3\cdot...\cdot10\cdot10}{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}\)
\(B=\dfrac{20}{11}\)
\(\Rightarrow11< 2x< 20\)
\(\Rightarrow x\in\left\{6;7;8;9\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a, Ta có:
\(\left(a+b+c\right)^2-\left(ab+bc+ca\right)=0\Leftrightarrow a^2+b^2+c^2+ab+bc+ca=0\)\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=0\Leftrightarrow a+b=b+c=c+a=0\)
\(\Leftrightarrow a=b=c=0\)
Vậy điều kiện để phân thức M được xác định là a, b, c không đồng thời = 0
b, Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
Đặt: \(a^2+b^2+c^2=x,ab+bc+ca=y\)
=> \(\left(a+b+c\right)^2=x+2y\)
Ta cũng có:
\(M=\dfrac{x\left(x+2y\right)+y^2}{x+2y-y}=\dfrac{x^2+2xy+y^2}{x+y}=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
\(=a^2+b^2+c^2+ab+bc+ca\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(sigma\frac{a^2+b^2}{ab\left(a+b\right)^3}\ge sigma\frac{\frac{\left(a+b\right)^2}{2}}{\left(a+b\right)^2\left(a^3+b^3\right)}=sigma\frac{1}{2\left(a^3+b^3\right)}\ge\frac{9}{4\left(a^3+b^3+c^3\right)}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
\(\Leftrightarrow a^4-2a^2b^2+b^4+a^2-2ac+c^2+a^2-2a+1\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(a-c\right)^2+\left(a-1\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}a=1\\b=\pm1\\c=1\end{matrix}\right.\)