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Áp dụng BĐT cosi ta có
\(\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\); \(\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge\frac{3}{b^2c}\); \(\frac{1}{c^3}+\frac{1}{c^3}+\frac{1}{d^3}\ge\frac{3}{c^2d}\)
\(\frac{1}{d^3}+\frac{1}{d^3}+\frac{1}{a^3}\ge\frac{3}{d^2a}\)
Cộng các BĐt trên ta có
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\ge\frac{1}{a^2b}+\frac{1}{b^2c}+\frac{1}{c^2d}+\frac{1}{d^2a}\)(1)
Áp dụng BĐT buniacoxki ta có
\(\left(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\right)\left(\frac{1}{a^2b}+\frac{1}{b^2c}+\frac{1}{c^2d}+\frac{1}{d^2a}\right)\ge \left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\right)^2\)
Kết hợp với (1) ta được ĐPCM
Dấu bằng xảy ra khi a=b=c
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
Bài 1 :
\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)
Bài 2 :
1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)
2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)
3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=\frac{1-\sqrt{3}}{5}\)
4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
Lời giải:
\(A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{48}\right)-\left(\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)\)
\(=2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{48}\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{24}-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)\)
\(=1-\left(\frac{1}{25}+\frac{1}{26}+...+\frac{1}{49}\right)\)
Chứng minh vế đầu:
Ta thấy:
\(\frac{1}{25}+\frac{1}{26}+...+\frac{1}{49}> \frac{1}{49}+\frac{1}{49}+...+\frac{1}{49}=\frac{25}{49}>\frac{25}{50}=\frac{1}{2}\)
\(\Rightarrow A=1-\left(\frac{1}{25}+\frac{1}{26}+...+\frac{1}{49}\right)< 1-\frac{1}{2}=\frac{1}{2}\) (đpcm)
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Vế sau sai, tính cụ thể thì $A< \frac{2}{5}$
mấy bài này thì bạn cứ đặt ẩn phụ cho dễ nhìn hơn mà giải nhé
a, \(\hept{\begin{cases}\frac{1}{2x-y}+x+3y=\frac{3}{2}\\\frac{4}{2x-y}-5\left(x+3y\right)=-3\end{cases}}\)ĐK : \(2x\ne y\)
Đặt \(\frac{1}{2x-y}=t;x+3y=u\)hệ phương trình tương đương
\(\hept{\begin{cases}t+u=\frac{3}{2}\\4t-5u=-3\end{cases}\Leftrightarrow\hept{\begin{cases}4t+4u=6\\4t-5u=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}9u=9\\4t=-3+5u\end{cases}}\Leftrightarrow\hept{\begin{cases}u=1\\t=\frac{-3+5}{4}=\frac{1}{2}\end{cases}}}\)
Theo cách đặt \(\hept{\begin{cases}x+3y=1\\\frac{1}{2x-y}=\frac{1}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x+3y=1\\2x-y=2\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+6y=2\\2x-y=2\end{cases}\Leftrightarrow}\hept{\begin{cases}7y=4\\x=\frac{y+2}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{4}{7}\\x=\frac{9}{7}\end{cases}}}\)
Vậy hệ pt có một nghiệm (x;y) = (9/7;4/7)
Khi \(n=1\to A=\frac{1}{5S_1^2}=\frac{5}{36}<\frac{35}{36}.\) Ta xét trường hợp \(n\ge2.\)
Theo giả thiết thì \(S_k=S_{k-1}+\frac{1}{5^k}>S_{k-1}\to S^2_k>S_k\cdot S_{k-1}\).
Vậy ta có \(\frac{1}{5^kS_k^2}<\frac{1}{5^kS_kS_{k-1}}=\frac{S_k-S_{k-1}}{S_kS_{k-1}}=\frac{1}{S_{k-1}}-\frac{1}{S_k}.\) Cho \(k=2,3,\ldots,n\) rồi cộng lại ta được
\(A<\frac{1}{5S_1^2}+\left(\frac{1}{S_1}-\frac{1}{S_2}\right)+\left(\frac{1}{S_2}-\frac{1}{S_3}\right)+\cdots+\left(\frac{1}{S_{n-1}}-\frac{1}{S_n}\right)\)
\(=\frac{1}{5S_1^2}+\frac{1}{S_1}-\frac{1}{S_n}<\frac{1}{5S_1^2}+\frac{1}{S_1}=\frac{5}{36}+\frac{5}{6}=\frac{35}{36}.\) (ĐPCM)