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

=> a+b=2c; b+c=2a; c+a=2b

Thay vào A ta được: A=((a+b)/b)((c+b)/c)((a+c)/a)

=2c/b.2a/c.2b/a=2.2.2=8

Bài 1: Ta có:

\(M=\frac{ad}{abcd+abd+ad+d}+\frac{bad}{bcd.ad+bc.ad+bad+ad}+\frac{c.abd}{cda.abd+cd.abd+cabd+abd}+\frac{d}{dab+da+d+1}\)

\(=\frac{ad}{1+abd+ad+d}+\frac{bad}{d+1+bad+ad}+\frac{1}{ad+d+1+abd}+\frac{d}{dab+da+d+1}\)

$=\frac{ad+abd+1+d}{ad+abd+1+d}=1$

Bài 2:

Vì $a,b,c,d\in [0;1]$ nên

\(N\leq \frac{a}{abcd+1}+\frac{b}{abcd+1}+\frac{c}{abcd+1}+\frac{d}{abcd+1}=\frac{a+b+c+d}{abcd+1}\)

Ta cũng có:
$(a-1)(b-1)\geq 0\Rightarrow a+b\leq ab+1$

Tương tự:

$c+d\leq cd+1$

$(ab-1)(cd-1)\geq 0\Rightarrow ab+cd\leq abcd+1$

Cộng 3 BĐT trên lại và thu gọn thì $a+b+c+d\leq abcd+3$

$\Rightarrow N\leq \frac{abcd+3}{abcd+1}=\frac{3(abcd+1)-2abcd}{abcd+1}$

$=3-\frac{2abcd}{abcd+1}\leq 3$

Vậy $N_{\max}=3$

3.

Hình vẽ:

Lời giải:

a) △AMC và △BNC có: \(\widehat{AMC}=\widehat{BNC}=90^0;\widehat{ACB}\) là góc chung.

\(\Rightarrow\)△AMC∼△BNC (g-g).

\(\Rightarrow\dfrac{AC}{BC}=\dfrac{CM}{CN}\Rightarrow AC.CN=BC.CM\left(1\right)\)

b) △AMB và △CPB có: \(\widehat{AMB}=\widehat{CPB}=90^0;\widehat{ABC}\) là góc chung.

\(\Rightarrow\)△AMB∼△CPB (g-g)

\(\Rightarrow\dfrac{AB}{CB}=\dfrac{BM}{BP}\Rightarrow AB.BP=BC.BM\left(2\right)\)

Từ (1) và (2) suy ra:

\(AC.CN+AB.BP=BC.CM+BC.BM=BC.\left(CM+BM\right)=BC.BC=BC^2\left(đpcm\right)\)b) Gọi \(M_0\) là trung điểm BC, giả sử \(AB< AC\).

\(\widehat{HBM}=90^0-\widehat{BHM}=90^0-\widehat{AHN}=\widehat{CAM}\)

△HBM và △CAM có: \(\widehat{HBM}=\widehat{CAM};\widehat{HMB}=\widehat{CMA}=90^0\)

\(\Rightarrow\)△HBM∼△CAM (g-g) 

\(\Rightarrow\dfrac{MH}{CM}=\dfrac{BM}{MA}\Rightarrow MH.MA=BM.CM\)

Ta có: \(BM.CM=\left(BM_0-MM_0\right)\left(CM_0+MM_0\right)=\left(BM_0-MM_0\right)\left(BM_0+MM_0\right)=BM_0^2-MM_0^2\le BM_0^2=\dfrac{BC^2}{4}\)

\(\Rightarrow MH.MA\le\dfrac{BC^2}{4}\).

Vì \(BC\) không đổi nên: \(max\left(MH.MA\right)=\dfrac{BC^2}{4}\), đạt được khi △ABC cân tại A hay A nằm trên đường trung trực của BC.

c) Sửa đề: \(S_1.S_2.S_3\le\dfrac{1}{64}.S^3\)

△AMC∼△BNC \(\Rightarrow\dfrac{AC}{BC}=\dfrac{MC}{NC}\Rightarrow\dfrac{AC}{MC}=\dfrac{BC}{NC}\)

△ABC và △MNC có: \(\dfrac{AC}{MC}=\dfrac{BC}{NC};\widehat{ACB}\) là góc chung.

\(\Rightarrow\)△ABC∼△MNC (c-g-c)

\(\Rightarrow\dfrac{S_{MNC}}{S_{ABC}}=\dfrac{S_1}{S}=\dfrac{MC}{AC}.\dfrac{NC}{BC}\left(1\right)\)

Tương tự: 

△ABC∼△MBP \(\Rightarrow\dfrac{S_{MBP}}{S_{ABC}}=\dfrac{S_2}{S}=\dfrac{MB}{AB}.\dfrac{BP}{BC}\left(2\right)\)

△ABC∼△ANP \(\Rightarrow\dfrac{S_{ANP}}{S_{ABC}}=\dfrac{S_3}{S}=\dfrac{AN}{AB}.\dfrac{AP}{AC}\left(3\right)\)

Từ (1), (2), (3) suy ra:

\(\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}=\left(\dfrac{MC}{AC}.\dfrac{NC}{BC}\right).\left(\dfrac{MB}{AB}.\dfrac{BP}{BC}\right).\left(\dfrac{AN}{AB}.\dfrac{AP}{AC}\right)\) 

\(\Rightarrow\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}=\left(\dfrac{MC.MB}{AC.AB}\right).\left(\dfrac{BP.AP}{AC.BC}\right).\left(\dfrac{AN.CN}{AB.BC}\right)\) (*)

Áp dụng câu b) ta có:

\(\left\{{}\begin{matrix}BM.CM\le\dfrac{1}{4}BC^2\\AP.BP\le\dfrac{1}{4}AB^2\\AN.CN\le\dfrac{1}{4}AC^2\end{matrix}\right.\)

Từ (*) suy ra:

\(\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}\le\left(\dfrac{\dfrac{1}{4}BC^2}{AC.AB}\right).\left(\dfrac{\dfrac{1}{4}AC^2}{AC.BC}\right).\left(\dfrac{\dfrac{1}{4}AB^2}{AB.BC}\right)=\dfrac{1}{64}\)

\(\Rightarrow S_1.S_2.S_3\le\dfrac{1}{64}.S^3\)

Dấu "=" xảy ra khi △ABC đều.

Áp dụng t/c dtsbn ta có:

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

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

\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)

Lời giải:

$\frac{2022a+b+c}{a}=\frac{a+2022b+c}{b}=\frac{a+b+2022c}{c}$

$=2021+\frac{a+b+c}{a}=2021+\frac{a+b+c}{b}=2021+\frac{a+b+c}{c}$

$\Rightarrow \frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$

$\Rightarrow a+b+c=0$ hoặc $\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$

$\Rightarrow a+b+c=0$ hoặc $a=b=c$

Nếu $a+b+c=0$ thì:

$P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{(-c)}{c}+\frac{(-b)}{b}+\frac{(-a)}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$ thì:

$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$

Áp dụng bất đẳng thức: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\) \(\Leftrightarrow a^2+2ab+b^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\left(đúng\right)\)

\(\dfrac{1}{2a+b+c}=\dfrac{1}{4}.\dfrac{4}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{2a}+\dfrac{1}{b+c}\right)\le\dfrac{1}{4}\left[\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\right]=\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c}\right)\)

CMTT \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\\\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\end{matrix}\right.\)

\(\Rightarrow M=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{2}{2a}+\dfrac{2}{2b}+\dfrac{2}{2c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}.4=1\)

\(minM=1\Leftrightarrow a=b=c=\dfrac{3}{4}\)

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

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

\(\Rightarrow P=2+2+2=6\)

                      Bài làm :

Ta có :

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(\Leftrightarrow\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)

\(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)

Dấu "=" xảy ra khi : a=b

Chứng minh tương tự như trên ; ta có :

\(\hept{\begin{cases}\frac{1}{b+c}\text{≤}\frac{1}{4}\left(\frac{1}{b}+\frac{1}{c}\right)\left(2\right)\\\frac{1}{c+a}\text{≤}\frac{1}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\left(3\right)\end{cases}}\)

Cộng vế với vế của (1) ; (2) ; (3) ; ta được :

\(A\text{≤}\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\text{=}\frac{3}{2}\)

Dấu "=" xảy ra khi ;

\(\hept{\begin{cases}a=b=c\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\end{cases}}\Leftrightarrow a=b=c=1\)

Vậy Max (A) = 3/2 khi a=b=c=1

quản lí tên kiểu j z

aaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffffaaaaaaaaaaaaaaaaaaaaaaffffffffffffffffffffffffffff

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

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

Do đó: 

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

Thay a+b=2c;b+c=2a và c+a=2b vào biểu thức \(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+b\right)}{abc}\), ta được: 

\(P=\dfrac{2a\cdot2b\cdot2c}{abc}=\dfrac{8abc}{abc}=8\)

Vậy: P=8

Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) = \(\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\) (t/c dãy tỉ số bằng nhau)

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

Ta cũng có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+b+c-a}{a+c}\) (t/c dãy tỉ số bằng nhau)

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

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{2b}{a+c}=1\) \(\Leftrightarrow\) a + c = 2b (*)

Tương tự ta cũng có: a + b = 2c (**); b + c = 2a (***)

Thay (*); (**); (***) vào P ta được:

P = \(\dfrac{2a.2b.2c}{abc}\) = 2.2.2 = 8

Vậy P = 8

Chúc bn học tốt!

Xét \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=126.16=2016\)

\(\Leftrightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=2016\)

\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=2013\)

Vậy A = 2013