\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.5.7.9+...+2^3.97.99.101}{1.3.5+3.5.7+5.7.9+...+97.99.101}\)

\(=\frac{2^3\left(1.3.5+3.5.7+5.7.9+...+97.99.101\right)}{1.3.5+3.5.7+5.7.9+...+97.99.101}=2^3=8\)

35/66

a)2.4/6.18=2.2.2/2.3.3.3.2=2/3.3.3=2/27

còn laj thì tương tự

\(A=\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+5.7.9+...+97.99.101}\)

\(A=\frac{8.\left(1.3.5+3.5.7+5.7.9+...+97.99.101\right)}{1.3.5+3.5.7+5.7.9+...+97.99.101}\)

\(A=8\)

\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Tất cả đều phải tìm điều kiện

\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)