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\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\left[c+d\ne0\right]\)

\(\left[a+b\right]\left[a+d\right]=\left[b+c\right]\left[c+d\right]\)

\(a^2+ad+ab+bd=bc+bd+c^2+cd\)

\(a^2+ad+ab=bc+c^2+cd\)

\(a^2+ad+ab-bc-c^2-cd=0\)

\(\left[a+c\right]\left[a-c\right]+d\left[a-c\right]+b\left[a-c\right]=0\)

\(\left[a-c\right]\left[a+b+c+d\right]=0\)

\(\orbr{\begin{cases}a-c=0\\a+b+c+d=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=c\\a+b+c+d=0\end{cases}}\)

bấm vào chỗ đúng đó nguyễn minh tâm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

\(\Rightarrow VT=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)

\(\Rightarrow VP=\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\left(2\right)\)

Từ (1) và (2) =>Đpcm

b, \(\frac{a+b}{a+b+c}>\frac{a+b}{a+b+c+d}\); \(\frac{b+c}{b+c+a}>\frac{b+c}{a+b+c+d}\)

 \(\frac{c+d}{c+d+a}>\frac{c+d}{a+b+c+d};\frac{d+a}{a+d+b}>\frac{a+d}{a+b+c+d}\)

Cộng các bĐT trên

=> \(B>\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

Ta  có Với \(0< \frac{x}{y}< 1\)

=> \(\frac{x}{y}< \frac{x+z}{y+z}\)

Áp dụng ta có 

\(B>\frac{a+b+d}{a+b+c+d}+...+\frac{d+a+c}{a+b+c+d}=3\)

Vậy 2<B<3

bạn áp dụng dãy tỉ số bằng nhau là xong

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

-->\(\frac{a}{b}=\frac{a-c}{b-d}\left(đpcm\right)\)

2) ta có \(\frac{a}{b}=\frac{c}{d}\)

đặt a=kb và c=kd

\(\frac{a+b}{a-b}=\frac{kb+b}{kb-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)

\(\frac{c+d}{c-d}=\frac{kd+d}{kd-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)

từ (1) và (2) --> \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)

Ta có :

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+c}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (1)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{3a^6+c^6}{3b^6+d^6}\)(2)

Từ (1) ; (2) \(\Rightarrow\frac{\left(a+c\right)^6}{\left(b+d\right)^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (đpcm)