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Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(a+b\right)\left(\dfrac{x^4}{a}+\dfrac{y^4}{b}\right)\ge\left(x^2+y^2\right)^2=1\)
\(\Rightarrow VT=\dfrac{x^4}{a}+\dfrac{y^4}{b}\ge\dfrac{1}{a+b}=VP\)
Dấu "=" khi \(\dfrac{x^2}{a}=\dfrac{y^2}{b}\)\(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\Rightarrow a+b=\dfrac{a}{x^2}\Rightarrow\left(a+b\right)^n=\dfrac{a^n}{x^{2n}}\)
Xét \(VT\) của biểu thức cần c.m:
\(VT=\left(\dfrac{x^2}{a}\right)^n+\left(\dfrac{y^2}{b}\right)^n=2\cdot\dfrac{x^{2n}}{a^n}\)
Và \(VP=\dfrac{2}{\left(a+b\right)^n}=\dfrac{2}{\dfrac{a^n}{x^{2n}}}=2\cdot\dfrac{x^{2n}}{a^n}\)
Vậy có ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ (d3)//(d4) \(\Leftrightarrow\left\{{}\begin{matrix}m^2+6m=7\\2n+7\ne-n^2-9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)\left(m+7\right)=0\\n^2+2n+16\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=1\\m=-7\end{matrix}\right.\\\left(n+1\right)^2+16\ne0\forall n\in R\end{matrix}\right.\)
KL: m=1 hoặc m=-7, \(n\in R\) thỏa mãn đề bài
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trước hết ta chứng minh BĐT
\(\frac{2k-1}{2k}< \frac{\sqrt{3k-2}}{\sqrt{3k+1}}\left(1\right)\)
Thật vậy, (1) \(\Leftrightarrow\left(2k-1\right)\sqrt{3k+1}< 2k\sqrt{3k-2}\)\(\Leftrightarrow\left(4k^2-4k+1\right)\left(3k+1\right)< 4k^2\left(3k-2\right)\)
\(\Leftrightarrow12k^3-8k^2-k+1< 12k^3-8k^2\)\(\Leftrightarrow k-1>0\left(\forall k\ge2\right)\)
Trong (1), lần lượt thay k bằng 1,2,...,n ta được:
\(\frac{1}{2}\le\frac{\sqrt{1}}{\sqrt{4}},\frac{3}{4}\le\frac{\sqrt{4}}{\sqrt{7}},....,\frac{2n-1}{2n}< \frac{\sqrt{3n-2}}{\sqrt{3n+1}}\)
Nhân từng vế các BĐT trên ta có:
\(\frac{1}{2}.\frac{3}{4}....\frac{2n-1}{2n}< \frac{\sqrt{1}}{\sqrt{4}}.\frac{\sqrt{4}}{\sqrt{7}}...\frac{\sqrt{3n-2}}{\sqrt{3n+1}}=\frac{1}{\sqrt{3n+1}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2. \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\)
\(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
3. Ta có: VT=\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}:\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{1-a}\right)=\left[\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}.\dfrac{1}{\sqrt{a}}\right].\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]=\dfrac{1+\sqrt{a}+a}{\sqrt{a}}.\dfrac{1}{1+\sqrt{a}}=\dfrac{1+\sqrt{a}+a}{\sqrt{a}+a}=\dfrac{1}{\sqrt{a}+a}+1\)
??? Sao rút gọn rồi ra kì vậy nhờ =="
1,
a.
\(\left[{}\begin{matrix}x-5\sqrt{x}+6\ne0\\\sqrt{x}-2\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\ne0\\\sqrt{x}\ne2\\\sqrt{x}\ne3\end{matrix}\right.\)
\(\left[{}\begin{matrix}\sqrt{x}\ne3\\\sqrt{x}\ne2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
Vậy ĐKXĐ : \(\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
\(A=7^{2n}-48n-1=\left(49^n-1\right)-48n=48\left[\left(49^{n-1}-1\right)+\left(49^{n-2}-1\right)+...+\left(49-1\right)\right]\)