Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b},\)ta có :

\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{b^2-bc+ac-a^2}{ab}\)

\(=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)

Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc},M.\frac{b}{c-a}=1+\frac{2b^3}{abc}.\)

Vậy \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=9\)

Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\)  (áp dụng Bất Đẳng Thức Cosi)

\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)

\(\Rightarrow\frac{a^2}{b}-a+2b\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\left(1\right)\)

Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)

Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)

Dấu "=" xảy ra khi a=b=c

1) Ta có : \(\frac{2016a+b+c+d}{a}=\frac{a+2016b+c+d}{b}=\frac{a+b+2016c+d}{c}=\frac{a+b+c+2016d}{d}\)

Trừ 4 vế với 2015 ta được : \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

Nếu a + b + c + d = 0

=> a + b = -(c + d)

=> b + c = (-a + d) 

=> c + d = -(a + b)

=> d + a = (-b + c)

Khi đó M = (-1) + (-1) + (-1) + (-1) = - 4

Nếu a + b + c + d\(\ne0\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\Rightarrow a=b=c=d\)

Khi đó M = 1 + 1 + 1 + 1 = 4

2) a) Ta có : \(\hept{\begin{cases}\left|x+2013\right|\ge0\forall x\\\left(3x-7\right)^{2004}\ge0\forall y\end{cases}\Rightarrow\left|x+2013\right|+\left(3x-7\right)^{2014}\ge0}\)

Dấu "=" xảy ra \(\hept{\begin{cases}x+2013=0\\3y-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2013\\y=\frac{7}{3}\end{cases}}}\)

b) 7^2x + 7^{2x + 3} = 344

=> 7^2x + 7^2x.7³ = 344

=> 7^2x.(1 + 7³) = 344

=> 7^2x  = 1

=> 7^2x = 7⁰

=> 2x = 0 => x = 0

c) Ta có :

 \(\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{5}{x+4}\Leftrightarrow\frac{7}{2x+2}=\frac{3}{2y-4}=\frac{10}{2x+8}=\frac{7-10}{2x+2-2x-8}=\frac{1}{2}\)(dãy tỉ số bằng nhau)

=>  2x + 2 = 14 => x = 6 ; 

2y - 4 = 6 => y = 5 ; 

6 + 5 + z = 17 => z = 6 

Vậy x = 6 ; y = 5 ; z = 6

3) a) Ta có : \(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)(dãy ti số bằng nhau) 

=> a + b + c = a + b - c => a + b + c - a - b + c = 0 => 2c = 0 => c = 0;  

Lại có : \(\frac{a+b+c}{a+b-c}-1=\frac{a-b+c}{a-b-c}-1\Leftrightarrow\frac{2c}{a+b-c}=\frac{2c}{a-b-c}\Rightarrow a+b-c=a-b-c\) => b = 0 

Vậy c = 0 hoặc b = 0

c) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b+b+c+a+c}{c+a+b}=2\)(dãy tỉ số bằng nhau) 

=> \(\hept{\begin{cases}a+b=2c\\b+c=2a\\a+c=2b\end{cases}}\)

Khi đó P = \(\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{b}{a}\right)=\frac{b+c}{b}.\frac{c+a}{c}=\frac{a+b}{a}=\frac{2a.2b.2c}{abc}=8\)

Vậy P = 8

2. b) \(7^{2x}+7^{2x+3}=344\)

        \(7^{2x}\cdot\left(1+7^3\right)=344\)

        \(7^{2x}\cdot\left(1+343\right)=344\)

        \(7^{2x}\cdot344=344\)

               \(7^{2x}=1\)  

               \(7^{2x}=7^0\)

              \(2x=0\)

               \(x=0\)

a)\(\left(x-\frac{1}{2}\right)^{2016},\left|\frac{3}{4}-y\right|\ge0\)

\(\left(x-\frac{1}{2}\right)^{2016}+\left|\frac{3}{4}-y\right|=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{2}\right)^{2016}=0\\\left|\frac{3}{4}-y\right|=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{4}-y=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{4}\end{cases}}\)

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

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

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

Ta có với a,b,c,d là các số thực khác 0 

\(\Rightarrow\frac{a-b+c+d}{b}=\frac{a+b-c+d}{c}=\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}\)

\(\Rightarrow\frac{a-b+c+d}{b}+1=\frac{a+b-c+d}{c}+1=\frac{a+b+c-d}{d}+1=\frac{b+c+d-a}{a}+1\)

\(\Rightarrow\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=\frac{b+c+d}{a}\)

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

\(\Rightarrow\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=\frac{b+c+d}{a}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)

Ta có M= \(\left(\frac{a+c+d}{b}\right)\left(\frac{a+b+d}{c}\right)\left(\frac{a+b+c}{d}\right)\left(\frac{b+c+d}{a}\right)\)

=> M= 3.3.3.3 

=> M =81

Áp dụng TC cuae DTSBN ta có:

a-b+c+d/b = a+b-c+d/c = a+b+c-d/d = b+c+d-a/a = \(\frac{a-b+c+d+a+b-c+d+a+b+c-d+b+c+d-a}{b+c+d+a}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)

=> a-b+c+d/b = 3 => a-b+c+d = 3b => a+c+d = 4b

a+b-c+d/c = 3 => a+b-c+d = 3c => a+b+d = 4c

a+b+c-d/d = 3 => a+b+c-d = 3d => a+b+c = 4d

b+c+d-a/a = 3 => b+c+d-a = 3a => b+c+d = 4a

=> M = \(\frac{\left(a+b+c\right)\left(a+b+d\right)\left(b+c+d\right)\left(c+d+a\right)}{abcd}=\frac{4d.4c.4a.4b}{abcd}=\frac{256abcd}{abcd}=256\)

Vậy M = 256

Theo đề ra, ta có: 

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

\(\rightarrow\frac{a+c+d}{b}=\frac{a+b+d}{c}=\frac{a+b+c}{d}=\frac{b+c+d}{a}=\frac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)

(Áp dụng dãy tỉ số bằng nhau)

Ta có:

\(M=\frac{\left(a+b+c\right)\left(a+b+d\right)\left(b+c+d\right)\left(c+d+a\right)}{abcd}=3^4=81\)

Mình chỉ làm bài 1a, và bài 3 thôi nhé,còn lại là bạn tự làm nhé

Bài 1:

a, Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

\(\Rightarrow\left[\frac{a}{b}\right]^2=\left[\frac{c}{d}\right]^2=\left[\frac{a+c}{b+d}\right]^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{(a+c)^2}{(b+d)^2}\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}\)

Bài 3 : Sửa đề : Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

CM : a = b = c

Cách 1 : Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)

vì \(a+b+c\ne0\)

\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c\)

Do đó : \(a=b=c\).

Cách 2 : Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=m\), ta có : \(a=bm,b=cm,c=am\)

Do đó : \(a=bm=m(mc)=m\left[m(ma)\right]\)

\(\Rightarrow a=m^3a\Rightarrow m^3=1(a\ne0)\Rightarrow m=1\)

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

Cách 3 : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}=\left[\frac{a}{b}\right]^3\Rightarrow1=\left[\frac{a}{b}\right]^3\Rightarrow\frac{a}{b}=1\)

Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)

a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

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

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

Cách 1 . \(A=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)

Đặt \(\frac{a-b}{c}=x\); \(\frac{b-c}{a}=y\) ; \(\frac{c-a}{b}=z\)

Ta có : \(\frac{x+y}{z}=\frac{\frac{a-b}{c}+\frac{b-c}{a}}{\frac{c-a}{b}}=\frac{ab\left(a-b\right)+cb\left(b-c\right)}{ac\left(c-a\right)}=\frac{b\left(b-a-c\right)}{ac}=\frac{2b^2}{ac}=\frac{2b^3}{abc}\)

tương tự : \(\frac{y+z}{x}=\frac{2c^3}{abc}\); \(\frac{x+z}{y}=\frac{2a^3}{abc}\)

\(\Rightarrow A=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)

\(=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)

Áp dụng bài toán phụ : Nếu a + b + c = 0 thì \(a^3+b^3+c^3=3abc\) (có thể chứng minh bằng cách rút a = - b - c  rồi thay vào tổng ba lập phương) được : 

\(A=3+\frac{2}{abc}.3abc=3+6=9\)

Đặt \(\frac{a-b}{c}=x=>\frac{c}{a-b}=\frac{1}{x}\)

\(\frac{b-c}{a}=y=>\frac{a}{b-c}=y\)

\(\frac{c-a}{b}=z=>\frac{b}{c-a}=\frac{1}{z}\)

=>\(A=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

=>\(A=x.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+y.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+z.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

=>\(A=1+\frac{x}{y}+\frac{x}{z}+1+\frac{y}{x}+\frac{y}{z}+1+\frac{z}{x}+\frac{z}{y}\)

=>\(A=3+\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\right)\)

=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}\)

Lại có: \(\frac{x+z}{y}=\frac{\frac{a-b}{c}+\frac{c-a}{b}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2}{bc}+\frac{c^2-ac}{bc}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2+c^2-ac}{bc}}{\frac{b-c}{a}}\)

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

\(=\frac{\left(a-b-c\right).\left(b-c\right).a}{\left(b-c\right).bc}=\frac{\left(a-b-c\right).a}{bc}=\frac{\left(a+a-a-b-c\right).a}{bc}\)

\(=\frac{\left[2a-\left(a+b+c\right)\right].a}{bc}\)

Vì a+b+c=0

=>\(\frac{x+z}{y}=\frac{\left(2a-0\right).a}{bc}=\frac{2a^2}{bc}=\frac{2a^3}{abc}\)

Chứng minh tương tự, ta có:

\(\frac{x+y}{z}=\frac{2b^3}{abc}\)

\(\frac{y+z}{x}=\frac{2c^3}{abc}\)

=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{3a^3}{abc}+\frac{3b^3}{abc}+\frac{3c^3}{abc}\)

=>\(A=3+\frac{2a^3+2b^3+2c^3}{abc}\)

=>\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}\)

Vì a+b+c=0

=>a=-(b+c)

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

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

=>\(a^3=-\left[b^3+3bc.\left(b+c\right)+c^3\right]\)

=>\(a^3=-b^3-3bc.\left(b+c\right)-c^3\)

=>\(a^3+b^3+c^3=-3bc.\left(b+c\right)\)

Vì a+b+c=0=>b+c=-a

=>\(a^3+b^3+c^3=-3bc.\left(-a\right)\)

=>\(a^3+b^3+c^3=3abc\)

Thay vào A, ta có:

\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+\frac{6.abc}{abc}=3+6=9\)

=>A=9

Vậy A=9

Cách 2. Đặt \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) ; \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)

\(\Rightarrow P=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)}{abc}\)

Xét riêng : \(ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)=ab\left[-\left(b-c\right)-\left(c-a\right)\right]+bc\left(b-c\right)+ac\left(c-a\right)\)

\(=\left[-ab\left(b-c\right)+bc\left(b-c\right)\right]+\left[-ab\left(c-a\right)+ac\left(c-a\right)\right]\)

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

Vậy : \(P=\frac{\left(c-a\right)\left(b-c\right)\left(b-a\right)}{abc}\)

Tiếp theo, rút gọn Q như sau : 

Đặt \(x=b-c\); \(y=c-a\); \(z=a-b\)

Ta có : \(x-y=a+b-2c=-c-2c=-3c\)

\(y-z=b+c-2a=-a-2a=-3a\)

\(z-x=c+a-2b=-b-2b=-3b\)

\(\Rightarrow3Q=\frac{-\left(y-z\right)}{x}+\frac{-\left(z-x\right)}{y}+\frac{-\left(x-y\right)}{z}\)\(\Rightarrow-3Q=\frac{y-z}{x}+\frac{z-x}{y}+\frac{x-y}{z}\)

Rút gọn tương tự như P, ta được : \(-3Q=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{xyz}=\frac{\left(-3c\right).\left(-3a\right).\left(3b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(\Rightarrow Q=-\frac{9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Vậy : \(A=PQ=\frac{\left(c-a\right)\left(c-b\right)\left(a-b\right)}{abc}.\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(\Rightarrow A=9\)