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TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(P=\dfrac{\left(b+c\right)}{b}.\dfrac{\left(a+b\right)}{a}.\dfrac{\left(a+c\right)}{c}=\dfrac{-a}{b}.\dfrac{-c}{a}.\dfrac{-b}{c}=-1\)
TH2: \(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a-b+c}{2b}=\dfrac{c-a+b}{2a}=\dfrac{a-c+b}{2c}=\dfrac{a-b+c+c-a+b+a-c+b}{2b+2a+2c}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b+c}{2b}=\dfrac{1}{2}\\\dfrac{c-a+b}{2a}=\dfrac{1}{2}\\\dfrac{a-c+b}{2c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+c=2b\\c+b=2a\\a+b=2c\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
a) A = 1/(1.2) + 1/(2.3) + ... + 1/[n(n + 1)]
= 1 - 1/2 + 1/2 - 1/3 + 1/n - 1/(n + 1)
= 1 - 1/(n + 1)
b) Do n ∈ ℕ
⇒ n + 1 > 0
⇒ 1/(n + 1) > 0
⇒ 1 - 1/(n + 1) < 1
Vậy A < 1
=>\(\dfrac{10a+b}{10b+c}=\dfrac{b}{c}\)
=>10ac+bc=10b^2+bc
=>ac=b^2
=>a/b=b/c=k
=>a=bk; b=ck
=>a=ck^2; b=ck
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{c^2k^4+c^2k^2}{c^2k^2+c^2}=k^2\)
\(\dfrac{a}{c}=\dfrac{ck^2}{c}=k^2\)
=>\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)
Vì \(\left|a\right|\le1;\left|b-1\right|\le2\)
\(=>\left|a\right|\cdot\left|b-1\right|=\left|ab-a\right|\le2\)
Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\) ta có:
\(\left|a-c+ab-a\right|\le\left|a-c\right|+\left|ab-a\right|=2+3=5\)
\(=>\left|ab-c\right|\le5\)
Tu \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
Hay \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Leftrightarrow a=b=c\)
Thay vao M ta co: \(M=\dfrac{a\cdot a+a\cdot a+a\cdot a}{a^2+a^2+a^2}=\dfrac{2019}{2019}=\dfrac{2018}{2018}=\dfrac{2017}{2017}=\dfrac{2016}{2015+1}=1\)
ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}\)
Áp dụng tính chất dãy tỉ sốbằng nhau ta có:
\(\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}=\dfrac{a^6+c^6}{b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
=\(\dfrac{c^6+3a^6}{d^6+3b^6}\)
\(\Rightarrow\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (ĐPCM)
Trước tiên, ta chứng minh \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với \(a,b>0\) (*)
(*) \(\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\), luôn đúng.
Vậy (*) được chứng minh. Dấu "=" xảy ra \(\Leftrightarrow a=b\)
\(\Rightarrow VT=a+b+\dfrac{1}{a}+\dfrac{1}{b}\ge a+b+\dfrac{4}{a+b}\)
Đặt \(a+b=t\left(0< t\le\dfrac{1}{2}\right)\)thì
\(VT\ge t+\dfrac{4}{t}\) \(=t+\dfrac{1}{4t}+\dfrac{15}{4t}\) (1)
Bây giờ ta sẽ chứng minh \(a+b\ge2\sqrt{ab}\) với \(a,b>0\) (**)
(**) \(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)
\(\Leftrightarrow\left(\sqrt{a}\right)^2-2\sqrt{a}\sqrt{b}+\left(\sqrt{b}\right)^2\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)
Vậy (**) được chứng minh. Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó từ (1) \(\Rightarrow VT\ge\left(t+\dfrac{1}{4t}\right)+\dfrac{15}{4t}\)
\(\ge2\sqrt{t.\dfrac{1}{4}t}+\dfrac{15}{4.\dfrac{1}{2}}\) (do \(0< t\le\dfrac{1}{2}\))
\(=\dfrac{17}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}t=a+b=\dfrac{1}{2}\\a=b\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{4}\)
Ta có đpcm.