Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

1)Xét \(VT=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2k^2+bdk^2}{d^2k^2-bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}=VP\)

Suy ra Đpcm

2)Xét \(VT=\frac{3\left(bk\right)^2+\left(dk\right)^2}{3b^2+d^2}=\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(1\right)\)

Xét \(VP=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(2\right)\)

Từ (1) và (2) suy ra Đpcm

a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)

Thay:

\(\frac{ab}{cd}=\frac{b^2}{d^2}\)

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

=> đpcm

Ta có: a/b = c/d => a/b.c/d = c/d.c/d (vì các p/s nào bằng nhau nhân với mấy cũng bằng nhau)

hay: ac/d = c^2/d^2 (1)

Lại có: a/b = c/d = a^2/b^2 = c^2/d^2 = a^2+c^2/b^2+d^2 (2)

Từ (1) và (2) => ac/bd = a^2+c^2/b^2/d^2

Ta có:\(\frac{3a+b+c+d}{a}=\frac{a+3b+c+d}{b}=\frac{a+b+3c+d}{c}=\frac{a+b+c+3d}{d}\)

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

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

\(TH1:a+b+c+d=0\Rightarrow\hept{\begin{cases}a+b=-\left(c+d\right)\\b+c=-\left(a+d\right)\end{cases}}\)

\(\Rightarrow Q=\left(\frac{-\left(c+d\right)}{c+d}\right)^2+\left(\frac{-\left(a+d\right)}{a+d}\right)^2+\left(\frac{c+d}{-\left(c+d\right)}\right)^2+\left(\frac{a+d}{-\left(a+d\right)}\right)^2\)

\(\Rightarrow Q=\left(-1\right)^2\cdot4=1\cdot4=4\)

\(TH2:a=b=c=d\)

\(\Rightarrow Q=\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2+\left(\frac{a+a}{a+a}\right)^2=1^2\cdot4=1\cdot4=4\)

Vậy Q=4

Gọi \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)(1)

Thay (1) vào ta có :

\(\frac{3a^2+c^2}{3b^2+d^2}=\frac{3\left(kb\right)^2+\left(kd\right)^2}{3b^2+d^2}=\frac{3k^2b^2+k^2+d^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\)(1)

\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(kb+kd\right)^2}{\left(b+d\right)^2}=\frac{\left[k\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\frac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)

Từ (1) và (2)

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

\(\RightarrowĐPCM\)

1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)

2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)

\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)

b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)