a: Sửa đề: \(\frac{a}{b}=\frac{a+c}{b+d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a}{b}=\frac{bk}{b}=k\)

\(\frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\)

Do đó: \(\frac{a}{b}=\frac{a+c}{b+d}\)

b: \(\frac{a-b}{a}=\frac{bk-b}{bk}=\frac{b\left(k-1\right)}{bk}=\frac{k-1}{k}\)

\(\frac{c-d}{c}=\frac{dk-d}{dk}=\frac{d\left(k-1\right)}{dk}=\frac{k-1}{k}\)

Do đó: \(\frac{a-b}{a}=\frac{c-d}{c}\)

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

\(\Leftrightarrow\frac{c+a-b}{\left(a-b\right)c}=\frac{a-b+c}{\left(b-c\right)a}\)(1)

Do \(\frac{a}{c}=\frac{a-b}{b-c}\Leftrightarrow a\left(b-c\right)=\left(a-b\right)c\)nên (1) đúng, đẳng thức được CM

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

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

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

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

\(=\frac{a}{2a}+\frac{2c}{c}\)

=1/2+2

=5/2

bt làm rồi ko cần giải nha 

v đăng lên làm j?:/

1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Tương tự :  \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)

\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)

a) We have :

       a² + b² + c² = ab + bc + ac 

<=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ac

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

<=> (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ac + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\)

b) We have :

a² - 2a + b² + 4b + 4c² - 4c + 6 = 0

(a² - 2a + 1) + (b² + 2.2b + 4) + (4c² - 4c + 1) = 0

(a - 1)² + (b + 2)² + (2c - 1)² = 0

\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b+2=0\\2c-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}\)