\(a,A=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,A=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}\)

Ta có \(a>0\Leftrightarrow\dfrac{1}{\sqrt{a}}>0\Leftrightarrow-\dfrac{1}{\sqrt{a}}< 0\Leftrightarrow A=1-\dfrac{1}{\sqrt{a}}< 1-0=1\left(đpcm\right)\)

\(c,A=\dfrac{1}{2}\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\left(tm\right)\)

\(a,=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ b,=x^2\left(x-2\right)-\left(x-2\right)\\ =\left(x-2\right)\left(x^2-1\right)=\left(x-1\right)\left(x-2\right)\left(x+2\right)\\ c,=\left(5-4x\right)\left(5+4x\right)\)

Kẻ DI,DJ lần lượt vuông góc với AK,CK

\(a,S_{AND}=\dfrac{1}{2}AN\cdot DI=\dfrac{1}{2}S_{ABCD}\) (chung đáy AD, cùng chiều cao hạ từ N)

\(b,S_{CDM}=\dfrac{1}{2}CM\cdot DJ=\dfrac{1}{2}S_{ABCD}\) (chung đáy CD, cùng chiều cao hạ từ M)

\(\Rightarrow\dfrac{1}{2}AN\cdot DI=\dfrac{1}{2}CM\cdot DJ\Rightarrow DI=DJ\left(AN=CM\right)\\ \Rightarrow\Delta DIK=\Delta DJG\left(ch-cgv\right)\\ \Rightarrow\widehat{IKD}=\widehat{JKD}\)

Vậy KD là phân giác \(\widehat{AKC}\)

Vì \(x_1\) là nghiệm PT nên \(x_1^2+3x_1-7=0\Leftrightarrow x_1^2=7-3x_1\)

\(F=x_1^2-3x_2-2013=7-3x_1-3x_2-2013\\ F=-3\left(x_1+x_2\right)-2006\)

Mà theo Viét ta có \(x_1+x_2=-3\)

\(\Rightarrow F=\left(-3\right)\left(-3\right)-2006=-1997\)

\(a,=2x^2+x-6-x^2-x=x^2-6\\ b,=5x^2-5y^2-x^2+y^2=4x^2-4y^2\\ c,=2-2x^2+3x^2-6x+x-2=x^2-5x\\ d,=4x^2+4xy-4x^2+4xy-y^2=8xy-y^2\\ e,=2a^2+2a-2a^2+1=2a+1\\ f,=4x^2+16x+16+9-4x^2=16x+25\\ g,=\left[x^2\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)=x^2+1\\ h,=\left(2x^4-6x^2+x^3-3x+x^2-3\right):\left(x^2-3\right)\\ =\left[2x^2\left(x^2-3\right)+x\left(x^2-3\right)+\left(x^2-3\right)\right]:\left(x^2-3\right)\\ =2x^2+x+1\)

\(a,=\dfrac{\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+5\right)-2\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\left(1-x\right)\\ =\dfrac{3x^2+8x-3-2x^2-3x+5-2x-2}{\left(x-1\right)\left(x+3\right)}\left(1-x\right)\\ =\dfrac{x\left(x+3\right)\left(1-x\right)}{x\left(x-1\right)\left(x+3\right)}=-x\)

\(b,=\dfrac{3x+9+x^2+x-2-x-7}{\left(x-1\right)\left(x+3\right)}\left(1-x\right)=\dfrac{x\left(x+3\right)\left(1-x\right)}{\left(x-1\right)\left(x+3\right)}=-x\)

\(c,=\dfrac{3x^2+x^2-1}{x\left(x-1\right)}\cdot\dfrac{x-1}{2x+1}=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x\left(x-1\right)}\cdot\dfrac{x-1}{2x+1}=\dfrac{2x-1}{x}\)

\(d,=\dfrac{3x^2-5x-2-x^2-2x+7x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2}\\ =\dfrac{2\left(x^2-1\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2}=x^2-1\)