\(4-x^2+2x=-\left(x^2-2x-4\right)=-\left(x^2-2x+1+3\right)\)

\(=-\left[\left(x-1\right)^2+3\right]=-\left(x-1\right)^2-3\le-3\)

Vậy GTLN của bt là -3\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(4x-x^2-5=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)\)

\(=-\left[\left(x-2\right)^2+1\right]=-\left(x-2\right)^2-1\le-1\)

Vậy GTLN của bt là -1\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right)n\left(n+1\right).4\)

         \(=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)\(-\left[\left(n-2\right)\left(n-1\right)n\left(n+1\right)\right]\)

          = \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)-0.1.2.3=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C+2A-3C}{2B+3D+2B-3D}=\frac{4A}{4B}=\frac{A}{B}\left(1\right)\)\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C-2A+3C}{2B+3D-2B+3D}=\frac{6C}{6D}=\frac{C}{D}\left(2\right)\)

Từ (1) và (2) suy ra : \(\frac{A}{B}=\frac{C}{D}\)

Giải :

Từ đảng thức : \(\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\)

\(\Rightarrow\left(2a+3c\right).\left(2b-3d\right)=\left(2b+3d\right).\left(2a-3c\right)\)

\(\Rightarrow4ab-6ad+6bc-9cd=4ab-6bc+6ad-9cd\)

\(\Rightarrow\left(4ab-6ad+6bc-9cd\right)-\left(4ab-6bc+6ad-9cd\right)=0\)

\(\Rightarrow4ab-6ad+6bc-9cd-4ab+6bc-6ad+9cd=0\)

\(\Rightarrow\left(4ab-4ab\right)-\left(6ad+6ad\right)+\left(6bc+6bc\right)-\left(9cd-9cd\right)=0\)

\(\Rightarrow-12ad+12bc=0\)

\(\Rightarrow12bc=12ad\)

\(\Rightarrow bc=ad\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)

\(\frac{2x-3}{-8}=\frac{512}{3-2x}\)

\(\left(2x-3\right)\cdot\left(3-2x\right)=-8\cdot512\)

\(\Rightarrow12x-4x^2-9=-4096\)

\(\Rightarrow-4x^2+12x-3^2=-4096\)

\(\Rightarrow-\left(4x^2-12x+3^2\right)=-4096\)

\(\Rightarrow-\left(2x-3\right)^2=-4096\)

\(\Rightarrow2x-3=4096\)

\(\Rightarrow2x=4099\)

\(\Rightarrow x=\frac{4099}{2}\)

Giải : 

Từ đẳng thức\(\frac{a}{a+c}=\frac{b}{b+d}\)

\(\Rightarrow a.\left(b+d\right)=\left(a+c\right).b\)

\(\Rightarrow ab+ad=ab+cb\)

\(\Rightarrow ad=cb\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)

Ta có:

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

Áp dụng tính chất dãy tỉ số bằng nhau:

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

=> \(\frac{a}{b}=\frac{c}{d}\).