ta có:

A+B=(a+b-5)+(-b-c+1)

      =a+b-5-b-c+1

      =a-c+(b-b)-(5-1)

      =a-c-4 (1)

Lại có:

C-D=(b-c-4)-(b-a)

     =b-c-4-b+a

     =(b-b)+a-c-4

     =a-c-4 (2)

Từ (1) và (2)=>A+B=C-D (vì cùng bằng a-c-4)

thieu de

Ta có:

\(A+B=a+b-5+\left(-b\right)-c+1\)

\(=a-c+\left(-b+b\right)+\left(-5+1\right)\)

\(=a-c-4\)

\(C-D=\left(b-c-4\right)-\left(b-a\right)\)

\(=b-c-4-b+a\)

\(=b-b+a-c-4\)

\(=a-c-4\)

Vậy: \(A+B=C-D\)

\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có : \(\frac{\left(a+c\right)^2}{a^2-c^2}=\frac{\left(a+c\right)^2}{a^2-ac+ac-c^2}=\frac{\left(a+c\right)^2}{a\left(a-c\right)+c\left(a-c\right)}=\frac{\left(a+c\right)^2}{\left(a+c\right)\left(a-c\right)}=\frac{a+c}{a-c}\)

\(=\frac{bk+dk}{bk-dk}=\frac{k\left(b+d\right)}{k\left(b-d\right)}=\frac{b+d}{b-d}\)(1)

Lại có \(\frac{\left(b+d\right)^2}{b^2-d^2}=\frac{\left(b+d\right)^2}{b^2-bd+bd-d^2}=\frac{\left(b+d\right)^2}{b\left(b-d\right)+d\left(b-d\right)}=\frac{\left(b+d\right)^2}{\left(b-d\right)\left(b+d\right)}=\frac{b+d}{b-d}\left(2\right)\)

Từ (1) (2) => \(\frac{\left(a+c\right)^2}{a^2-c^2}=\frac{\left(b+d\right)^2}{b^2-d^2}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(\frac{\left(a+c\right)^2}{a^2-c^2}=\frac{\left(a+c\right)\left(a+c\right)}{\left(a-c\right)\left(a+c\right)}=\frac{a+c}{a-c}=\frac{bk+dk}{bk-dk}=\frac{k\left(b+d\right)}{k\left(b-d\right)}=\frac{b+d}{b-d}\)(1)

\(\frac{\left(b+d\right)^2}{b^2-d^2}=\frac{\left(b+d\right)\left(b+d\right)}{\left(b-d\right)\left(b+d\right)}=\frac{b+d}{b-d}\)(2)

Từ (1) và (2) => đpcm 

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

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

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

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

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

\(\Leftrightarrow\frac{\left(a+c\right)^2}{a^2-c^2}=\frac{\left(b+d\right)^2}{b^2-d^2}\) ( đpcm )

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2a}{2c}=\frac{b}{d}\)

áp dụng tính chất của dãy tỉ số bằng nhau ta có

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

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

con ai bit ko