Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)

a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)\(=\dfrac{\dfrac{a}{k}.b}{\dfrac{c}{k}.d}=\dfrac{ab}{cd}=VT\)

Vậy...

b) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)

\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)

Suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)

c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\left(bk\right)^2+3\left(bk\right).b}{11\left(bk\right)^2-8b^2}\)\(=\dfrac{7k^2+3k}{11k^2-8}\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3\left(dk\right).d}{11\left(dk\right)^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)

Suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

a) Đk: \(x\ge9;x\ne13\)

\(P=\dfrac{x-9-4}{\sqrt{x-9}-2}=\dfrac{\left(\sqrt{x-9}-2\right)\left(\sqrt{x-9}+2\right)}{\sqrt{x-9}-2}=\sqrt{x-9}+2\)

b) \(P=\sqrt{x-9}+2\ge2\)

Dấu "="xảy ra \(\Leftrightarrow x=9\)

Vậy GTNN của P là 2

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c+2a-3c}{2b+3d+2b-3d}=\dfrac{a}{b}\)

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c-\left(2a-3c\right)}{2b+3d-\left(2b-3d\right)}=\dfrac{c}{d}\)

Suy ra \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(B=\sqrt{\dfrac{a+6}{a+1}}\) ( ĐK: \(a>-1;a\le-6\) )

\(\Rightarrow B^2=\dfrac{a+6}{a+1}=1+\dfrac{5}{a+1}\)

Với \(B\in Z\Rightarrow B^2\in Z\Leftrightarrow\dfrac{5}{a+1}\in Z\) 

a) mà \(a\in Z\) nên \(a+1\inƯ\left(5\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a+1=\pm1\\a+1=\pm5\end{matrix}\right.\)\(\Leftrightarrow a=0\) ,\(a=4\) hoặc \(a=-6\)

Tại \(a=0\Leftrightarrow B=\sqrt{6}\) (loại)

Tại \(a=4\Rightarrow B=\sqrt{2}\) (loại)

Tại \(a=-6\Rightarrow B=0\) (tm)

Vậy \(a=-6\)

b) Thay \(a=\dfrac{4}{9}\Rightarrow B=\dfrac{\sqrt{754}}{13}\) 
Hm...
c) Đợi cao nhân. Đề này quá sức của thần.

Xét \(\Delta'=\left(m+1\right)^2-\left(-m-4\right)=m^2+3m+5=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}>0\forall m\)

Suy ra pt có hai nghiệm pb với mọi m

Theo hệ thức viet có:

\(\left\{{}\begin{matrix}x_1+x_2=-2m-2\\x_1x_2=-m-4\end{matrix}\right.\)

\(M=x_1-x_1^2+x_2-x_2^2=x_1+x_2-\left(x_1+x_2\right)^2+2x_1x_2\)

\(=-2m-2-\left(-2m-2\right)^2+2\left(-m-4\right)\)

Qua đó thấy M phụ thuộc vào m

Thề luôn ngươi trình bày xấu vchưởng ahuhu 

\(\sqrt{x}-2+7=\sqrt{x}+5\) đó

\(Q=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\dfrac{\sqrt{x}-1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-2\right)\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)

Đk: \(x\ge0,x\ne4\)

\(Q=\dfrac{\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-2\right)\)

\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)

k) Sửa chút:
Cách đều khi \(\left|a\right|=\left|b\right|\)

\(\Rightarrow a=b\) hoặc \(a=-b\) kết hợp với \(b=\dfrac{1}{7}a+\dfrac{3}{7}\)

Vậy \(I\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) hoặc \(I\left(-\dfrac{3}{8};\dfrac{3}{8}\right)\)