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:

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

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

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

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

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

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

rêtrt

A= 5/4

B= 63/52

mấy cái đó từ công thức mà ra

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)

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

Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

b: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)

\(\left(\dfrac{a-b}{c-d}\right)^2=\left(\dfrac{bk-b}{dk-d}\right)^2=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)

a)

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

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

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

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

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

\(C=\dfrac{\left(b-c+c-a\right)^3+3\left(b-c\right)\left(c-a\right)\left(b-c+c-a\right)+\left(a-b\right)^3}{a^2b-a^2c+b^2c-b^2a+c^2a-c^2b}\)

\(=\dfrac{3\left(b-c\right)\left(c-a\right)\left(b-a\right)}{a^2b-b^2a-a^2c+b^2c+c^2a-c^2b}\)

\(=\dfrac{3\left(b-c\right)\left(c-a\right)\left(b-a\right)}{\left(a-b\right)\cdot ab-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}\)

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

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

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

Bài 1:

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

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

\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)

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

CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)

\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Bài 2:

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

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

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

\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)

\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)

\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

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

\(=2\left(a+b+c\right)\)