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AM-GM :\(\dfrac{1}{a^4+b^2+2ab^2}=\dfrac{1}{a^4+b^2+ab^2+ab^2}\le\dfrac{1}{4\sqrt[4]{a^6b^6}}\)
\(\Rightarrow Q\le\dfrac{1}{2\sqrt[4]{a^6b^6}}\) (1)
AM - GM : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\Leftrightarrow2\ge\dfrac{2}{\sqrt{ab}}\Leftrightarrow ab\ge1\) (2)
Kết hợp (1) và (2) ta có đpcm
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\)
Đặt \(A=\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
\(=\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
\(=n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
\(=\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}=n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2}\right)=n.B\)
\(A:B=n\)
Ta có: \(1-\dfrac{1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)
Thế vô bài toán ta được
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{n^2}\right)=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{\left(n-1\right)\left(n+1\right)}{n.n}=\dfrac{1}{2}.\dfrac{n+1}{n}\)
Ta thấy
\(\dfrac{1}{2}.\dfrac{n}{n}< \dfrac{1}{2}.\dfrac{n+1}{n}< \dfrac{1}{2}.\dfrac{n+n}{n}\)
\(\Rightarrow\dfrac{1}{2}< \dfrac{1}{2}.\dfrac{n+1}{n}< 1\)
\(\Rightarrow\)ĐPCM
b.
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)
bài 1:
a) 4n+4+3n-6<19
<=> 7n-2<19
<=> 7n<21 <=> n< 3
b) n\(^2\) - 6n + 9 - n\(^2\) + 16\(\leq\)43
-6n+25\(\leq\)43
-6n\(\leq\)18
n\(\geq\)-3
1 ) \(A=\left(\dfrac{2x^3+2}{x+1}-2x\right)\left(\dfrac{x^3-1}{x-1}+x\right)\)
\(\Leftrightarrow A=\left(\dfrac{2x^3+2-2x^2-2x}{x+1}\right)\left(x^2+2x+1\right)\)
\(\Leftrightarrow A=\left(\dfrac{\left(2x^2-2\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=\left(\dfrac{2\left(x-1\right)\left(x+1\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=2\left(x-1\right)^2\left(x+1\right)^2\ge0\forall x\)
Ta có : a-\(\dfrac{1}{a}-2=a^2-2a+1=\left(a-1\right)^2\ge0\)
\(\Rightarrow a-\dfrac{1}{a}\ge2\)
Q(x)=2x2+\(\dfrac{2}{x^2}+3y^2+\dfrac{3}{y^2}+\dfrac{4}{x^2}+\dfrac{5}{y^2}\)
=2(\(x^2+\dfrac{1}{x^2}\)) +3(\(y^2+\dfrac{1}{y^2}\))+(\(\dfrac{4}{x^2}+\dfrac{5}{y^2}\))
\(\ge2.2+3.2+9=19\)
Dấu = xảy ra khi x=y=1
Em chưa học làm dạng này , em làm thử thôi nhá, sai xin chỉ dạy thêm nha
2 . \(\dfrac{n^7+n^2+1}{n^8+n+1}=\dfrac{n^7-n+n^2+n+1}{n^8-n^2+n^2+n+1}\)
\(=\dfrac{n\left(n^6-1\right)+n^2+n+1}{n^2\left(n^6-1\right)+n^2+n+1}=\dfrac{n\left(n^3+1\right)\left(n^3-1\right)+n^2+n+1}{n^2\left(n^3+1\right)\left(n^3-1\right)+n^2+n+1}\)\(=\dfrac{n\left(n^3+1\right)\left(n-1\right)\left(n^2+n+1\right)+n^2+n+1}{n^2\left(n^3+1\right)\left(n-1\right)\left(n^2+n+1\right)+n^2+n+1}\)
\(=\dfrac{\left(n^2+n+1\right)\left[\left(n^4+n\right)\left(n-1\right)\right]}{\left(n^2+n+1\right)\left[\left(n^5+n^2\right)\left(n-1\right)+1\right]}\)
\(=\dfrac{n^5-n^4+n^2-n}{n^6-n^5+n^3-n^2+1}=\dfrac{n^4\left(n-1\right)+n\left(n-1\right)}{n^5\left(n-1\right)+n^2\left(n-1\right)+1}\)
\(=\dfrac{\left(n-1\right)\left(n^4+n\right)}{\left(n-1\right)\left(n^5+n^2\right)+1}\)
Vậy ,với mọi số nguyên dương n thì phân thức trên sẽ không tối giản
- Áp dụng BĐT cauchuy : \(\left\{{}\begin{matrix}9m^2+n^2\ge2\sqrt{9m^2n^2}=6mn\\\dfrac{1}{9m^2}+\dfrac{1}{n^2}\ge2\sqrt{\dfrac{1}{9m^2n^2}}=\dfrac{2}{3mn}\end{matrix}\right.\)
\(\Rightarrow\left(9m^2+n^2\right)\left(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\right)\ge6mn.\dfrac{2}{3mn}=4\left(1\right)\)
- Dấu " = " xảy ra <=> \(9m^2=n^2\)\(\Leftrightarrow\left(3m-n\right)\left(3m+n\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=3m\\n=-3m\end{matrix}\right.\)
Mà m, n > 0
\(\Rightarrow n=3m\)
- Không biết đề bài là gì ??
dạ chứng minh pt này lớn hơn hoặc bằng 4 ạ
Tìm ẩn n và m
Là 9m2 + n2.(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) hay là (9m2 + n2)(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) ?
A nghĩ đề là (9m2 + n2)(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) \(\ge\) 4
Cách khác:
Ta có BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) (a; b dương)
CM: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\) \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\) (a + b)2 \(\ge\) 4ab
\(\Leftrightarrow\) (a + b)2 - 4ab \(\ge\) 0
\(\Leftrightarrow\) (a - b)2 \(\ge\) 0 (luôn đúng)
Dấu "=" xảy ra \(\Leftrightarrow\) a = b
Áp dụng BĐT phụ trên cho 2 số \(\dfrac{1}{9m^2}\) và \(\dfrac{1}{n^2}\) dương ta được:
\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\ge\dfrac{4}{9m^2+n^2}\)
\(\Leftrightarrow\) (9m2 + n2)(\(\dfrac{1}{9m^2}+\dfrac{1}{n^2}\)) \(\ge\) \(\dfrac{4\left(9m^2+n^2\right)}{9m^2+n^2}\) = 4 (đpcm)Dấu "=" xảy ra \(\Leftrightarrow\) 9m2 = n2
\(\Leftrightarrow\) n = 3m (do m, n > 0)
\(\Leftrightarrow\) m = \(\dfrac{n}{3}\)
Vậy ...