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\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{2450}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{7}-...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+..+\dfrac{1}{50}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(A=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=B\)
\(\Rightarrow A:B=1\)
Coi như a, b, c là số dương
Áp dụng BĐT Cô-si ta có:
\(\dfrac{a}{bc}+\dfrac{c}{ba}\ge2\sqrt{\dfrac{a}{bc}.\dfrac{c}{ba}}=2\sqrt{\dfrac{1}{b^2}}=\dfrac{2}{b}\left(1\right)\)
Dấu "=" xảy ra ...
\(\dfrac{a}{bc}+\dfrac{b}{ac}\ge2\sqrt{\dfrac{a}{bc}.\dfrac{b}{ac}}=2\sqrt{\dfrac{1}{c^2}}=\dfrac{2}{c}\left(2\right)\)
Dấu "=" xảy ra ...
\(\dfrac{c}{ba}+\dfrac{b}{ac}\ge2\sqrt{\dfrac{c}{ba}+\dfrac{b}{ac}}=2\sqrt{\dfrac{1}{a^2}}=\dfrac{2}{a}\left(3\right)\)
Dấu "=" xảy ra ...
Từ (1), (2), (3) ta có:
\(\dfrac{a}{bc}+\dfrac{c}{ba}+\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ba}+\dfrac{b}{ac}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\\ \Rightarrow2\left(\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ba}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\\ \Rightarrow\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ba}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Dấu "=" xảy ra ...
Vậy ...
a) A = {\(\dfrac{1}{n\left(n+1\right)}\)| \(n\in\mathbb{N},1\le n\le5\)}
b) B = {\(\dfrac{1}{n^2-1}\)|\(n\in\mathbb{N},2\le n\le6\)\(\)}
Áp dụng BĐT cosi:
\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)
Dấu \("="\Leftrightarrow a=b=c\)
1: \(\Leftrightarrow a\sqrt{a}+b\sqrt{b}>=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\)
=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b-\sqrt{ab}\right)>=0\)
=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2>=0\)(luôn đúng)
a) Ta có: \(1-\frac{\sin^2x}{1+\cot x}-\frac{\cos^2x}{1+\tan x}=1-\frac{\sin^2x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2x}{1+\frac{\sin x}{\cos x}}\) (Đk: sinx và cosx khác 0)
\(=1-\frac{\sin^3x}{\sin x+\cos x}-\frac{\cos^3x}{\cos x+\sin x}\)
\(=1-\frac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x.\cos x+\cos^2x\right)}{\sin x+\cos x}\)
\(=1-\left(\sin^2x+\cos^2x-\sin x.\cos x\right)\) (do sinx + cosx luôn khác 0)
\(=\sin x.\cos x\) ( do \(\sin^2x+\cos^2x=1\))
b) Ta có: \(\frac{\sin^2x+2\cos x-1}{2+\cos x-\cos^2x}=\frac{\left(\sin^2x-1\right)+2\cos x}{-\left(\cos x+1\right)\left(\cos x-2\right)}\) (Đk: cosx khác -1 và 2)
\(=\frac{-\cos x\left(\cos x-2\right)}{-\left(\cos x+1\right)\left(\cos x-2\right)}\)
\(=\frac{\cos x}{1+\cos x}\)
a) Ta có: 1−sin2x1+cotx −cos2x1+tanx =1−sin2x1+cosxsinx −cos2x1+sinxcosx (Đk: sinx và cosx khác 0)
=1−sin3xsinx+cosx −cos3xcosx+sinx
=1−(sinx+cosx)(sin2x−sinx.cosx+cos2x)sinx+cosx
=1−(sin2x+cos2x−sinx.cosx) (do sinx + cosx luôn khác 0)
=sinx.cosx ( do sin2x+cos2x=1)
b) Ta có: sin2x+2cosx−12+cosx−cos2x =(sin2x−1)+2cosx−(cosx+1)(cosx−2) (Đk: cosx khác -1 và 2)
=−cosx(cosx−2)−(cosx+1)(cosx−2)
=cosx1+cosx
\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\)
\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)
Dự đoán điểm rơi xảy ra tại \(\left(a;b;c\right)=\left(3;2;4\right)\)
Đơn giản là kiên nhẫn tính toán và tách biểu thức:
\(D=13\left(\dfrac{a}{18}+\dfrac{c}{24}\right)+13\left(\dfrac{b}{24}+\dfrac{c}{48}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{2}{ab}\right)+\left(\dfrac{a}{18}+\dfrac{c}{24}+\dfrac{2}{ac}\right)+\left(\dfrac{b}{8}+\dfrac{c}{16}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{b}{6}+\dfrac{c}{12}+\dfrac{8}{abc}\right)\)
Sau đó Cô-si cho từng ngoặc là được
