Ta có: \(A\cdot C+B^2-2x^4y^4=x^3y\cdot xy^3+\left(x^2y^2\right)^2-2x^4y^4\)

\(\Leftrightarrow A\cdot C+B^2-2x^4y^4=x^4y^4+x^4y^4-2xy^4\)

\(\Leftrightarrow A\cdot C+B^2-2x^4y^4=0\)(đpcm)

A.C + B^2 - 2x^4.y^4

=(x^3.y)(x.y^3) + x^4.y^4 - 2x^4.y^4

=(x^4.y^4 + x^4.y^4) - 2x^4.y^4

=2x^4.y^4 - 2x^4.y^4

=0

Từ \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\left(1\right)\)

\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\left(2\right)\)

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

Từ (1) (2) (3) \(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)

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

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

\(\Rightarrow a=b=c=d\)

Khi đó M = \(\frac{a}{b+c+d}+\frac{b}{a+c+d}=\frac{a}{3a}+\frac{a}{3a}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)

Vậy \(M=\frac{2}{3}\)

?

\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\left(đpcm\right)\)

Thay b² = ac vào biểu thức trên, ta có:

\(\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)

\(\Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)

\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\left(1\right)\)

Ta có \(b^2=ac\Leftrightarrow\dfrac{ac}{c^2}=\dfrac{b^2}{c^2}\Leftrightarrow\dfrac{a}{c}=\dfrac{b^2}{c^2}\left(2\right)\)

\(\left(1\right)\left(2\right)\LeftrightarrowĐpcm\)