Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b-c}{3c}=\dfrac{b+c-a}{3a}=\dfrac{c+a-b}{3b}=\dfrac{a+b-c+b+c-a+c+a-b}{3a+3b+3c}=\dfrac{a+b+c+\left(a-a\right)+\left(b-b\right)+\left(c-c\right)}{3a+3b+3c}=\dfrac{a+b+c}{3\left(a+b+c\right)}=\dfrac{1}{3}\)

Khi đó:

\(\left\{{}\begin{matrix}\dfrac{a+b-c}{3c}=\dfrac{1}{3}\\\dfrac{b+c-a}{3a}=\dfrac{1}{3}\\\dfrac{c+a-b}{3b}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b-3c=3c\\3b+3c-3a=3a\\3c+3a-3b=3b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6c\\3b+3c=6a\\3c+3a=6b\end{matrix}\right.\)Thay vào \(P\)

\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\left(\dfrac{a+b}{a}\right)\left(\dfrac{c+a}{c}\right)\left(\dfrac{b+c}{b}\right)\)

\(27P=3\left(\dfrac{a+b}{a}\right).3\left(\dfrac{c+a}{c}\right).3\left(\dfrac{b+c}{b}\right)\)

\(27P=\left(\dfrac{3a+3b}{a}\right)\left(\dfrac{3c+3a}{c}\right)\left(\dfrac{3b+3c}{b}\right)\)

\(27P=\)\(\dfrac{6c}{a}.\dfrac{6b}{c}.\dfrac{6a}{b}=\dfrac{216abc}{abc}=216\Leftrightarrow P=\dfrac{216}{27}=8\)

Thanks bạn nhiều

@Akai Haruma

Từ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)

\(\Leftrightarrow a^2b+abc+a^2c+b^2a+b^2c+abc+bc^2+ac^2=0\)

\(\Leftrightarrow ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)

\(\Leftrightarrow\left(ab+ac+bc+c^2\right)\left(a+b\right)=0\)

\(\Leftrightarrow\left[a\left(b+c\right)+c\left(b+c\right)\right]\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+c\right)\left(b+c\right)\left(a+b\right)=0\)

Thay vào từng TH suy ra M=0

a)

\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)

\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)

\(=\dfrac{1}{100}.\dfrac{101}{2}\)

\(=\dfrac{101}{200}\)

Câu 1: 

a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)

\(=a^3+b^3\)

b: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

C₁: +Ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}\)\(=\dfrac{2a+2b+2c}{a+b+c}\)

\(=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)

\(=\left(\dfrac{b}{b}+\dfrac{a}{b}\right)\cdot\left(\dfrac{c}{c}+\dfrac{b}{c}\right)\cdot\left(\dfrac{a}{a}+\dfrac{c}{a}\right)\)

\(=\dfrac{b+a}{b}\cdot\dfrac{c+b}{c}\cdot\dfrac{a+c}{a}\)\(=\dfrac{b+a}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}\)\(=2\cdot2\cdot2=8\)

\(\Rightarrow M=8\)

C₂:

+)Ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)

\(\Rightarrow\dfrac{a+b}{c}+1=\dfrac{b+c}{a}+1=\dfrac{c+a}{b}+1\)

\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)

+)Vậy, ta có:

\(\dfrac{a+b+c}{c}=\dfrac{b+c+a}{c}\) và \(\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)

\(-\)Vì \(\dfrac{a+b+c}{c}=\dfrac{b+c+a}{c}\)

\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{b+c+a}{a}=0\)

\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\) ₍₁₎

\(-\)Vì \(\dfrac{b+c+a}{a}=\dfrac{c+a+b}{b}\)

\(\Rightarrow\dfrac{b+c+a}{a}-\dfrac{c+a+b}{b}=0\)

\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\) ₍₂₎

\(-\)Mà theo đề bài ta có a,b,c đôi một khác 0 nên:

Từ _{(1) và(2)} ta suy ra được:

\(\rightarrow\)a+b+c=0

\(\Rightarrow a+b=\left(-c\right)\)

\(\Rightarrow b+c=\left(-a\right)\)

\(\Rightarrow c+a=\left(-b\right)\)

+)Ta có:

M= \(\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)

=\(\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{c}\)

\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{c}=\left(-1\right)\cdot\left(-1\right)\cdot\left(-1\right)\) =(-1)

Vậy M= \(\left\{{}\begin{matrix}8\\-1\end{matrix}\right.\)

Làm lại cho you đây -_- vừa nãy bấm mt nhầm,đời t nhọ vãi

1)\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+....+16\right)\)

\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+\dfrac{1+2+3+4}{4}+...+\dfrac{1+2+3+...+16}{16}\)

Xét thừa số tổng quát: \(\dfrac{1+2+3+...+t}{t}=\dfrac{\left[\left(t-1\right):1+1\right]:2.\left(t+1\right)}{t}=\dfrac{\dfrac{t}{2}\left(t+1\right)}{t}=\dfrac{\dfrac{t^2}{2}+\dfrac{t}{2}}{t}=\dfrac{t\left(\dfrac{t}{2}+\dfrac{1}{2}\right)}{t}=\dfrac{t}{2}+\dfrac{1}{2}\)

Như vậy: \(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+....+\dfrac{17}{2}\)

\(P=\dfrac{2+3+4+5+...+17}{2}\)

\(P=\dfrac{152}{2}=76\)

2) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)

\(\Rightarrow2016\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{2016}{a+b}+\dfrac{2016}{b+c}+\dfrac{2016}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2016}{3}-1-1-1=\dfrac{2007}{3}\)

\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)

Xét thừa số tổng quát: \(\dfrac{1+2+3+..+n}{n}=\dfrac{\left[\left(n-1\right):1+1\right]:2.\left(n+1\right)}{n}=\dfrac{\dfrac{n}{2}\left(n+1\right)}{n}=\dfrac{\dfrac{n^2}{2}+\dfrac{n}{2}}{n}=\dfrac{n\left(\dfrac{n}{2}+\dfrac{1}{2}\right)}{n}=\dfrac{n}{2}+\dfrac{1}{2}\)

Như vậy:

\(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{2}{2}+\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{1}{2}+\dfrac{4}{2}+\dfrac{1}{2}+...+\dfrac{16}{2}+\dfrac{1}{2}\)

\(P=1+\left(\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{16}{2}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+...+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{2+3+4+...+16}{2}+\dfrac{15}{2}\)

\(P=1+\dfrac{\left[\left(16-2\right):1+1\right]:2.\left(16+2\right)}{2}+\dfrac{15}{2}\)

\(P=1+210+\dfrac{15}{2}=218,5\)

P(x)=\(ax^2+bx+c\) (1)(a\(\ne0\) )

Ta có : 

\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}\)\(\Rightarrow\left\{{}\begin{matrix}b=2a\\c=3a\end{matrix}\right.\)(2)

Thay(2) vào (1)\(\Rightarrow P\left(x\right)=ax^2+2ax+3a\)

\(\Rightarrow\dfrac{P\left(-2\right)-3P\left(-1\right)}{a}=\dfrac{4a-4a+3a-3\left(a-2a+3a\right)}{a}\)=\(\dfrac{3a-3a+6a-9a}{a}=\dfrac{-3a}{a}=-3\)

Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\c+a=-b\\a+b=-c\end{matrix}\right.\)

\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{-abc}{abc}=-1\)

Với \(a+b+c\ne0\)

\(\dfrac{a+b-2021c}{c}=\dfrac{b+c-2021a}{a}=\dfrac{c+a-2021b}{b}=\dfrac{-2019\left(a+b+c\right)}{a+b+c}=-2019\\ \Leftrightarrow\left\{{}\begin{matrix}a+b-2021c=-2019c\\b+c-2021a=-2019a\\c+a-2021b=-2019b\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

\(B=\dfrac{a+b}{a}\cdot\dfrac{a+c}{c}\cdot\dfrac{b+c}{b}=\dfrac{2a\cdot2b\cdot2c}{abc}=8\)

Với a+b+c=0⇔⎧⎪⎨⎪⎩b+c=−ac+a=−ba+b=−ca+b+c=0⇔{b+c=−ac+a=−ba+b=−c