Bài 7:

a: \(A=x+\sqrt{x}\ge0\forall x\)

Dấu '=' xảy ra khi x=0

\(a,P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x} +1\right)}\)

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

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(---\)

\(b,P< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\left(vì.2\left(\sqrt{x}+1\right)>0\forall x\ge0\right)\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Kết hợp với điều kiện của \(x\), ta được:

\(0\le x< 9;x\ne1\) thì \(P< \dfrac{1}{2}\)

#\(Toru\)

Bài 1:

1) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

2) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

3) \(\Rightarrow\left(4x-3\right)\left(7-12x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{7}{12}\end{matrix}\right.\)

4) \(\Rightarrow x^3+8-x^3+25x=-17\)

\(\Rightarrow25x=-25\Rightarrow x=-1\)

5) \(\Rightarrow\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)

\(\Rightarrow\left(3x-2\right)\left(3x+2-6x+4\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(-3x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

Bài 3: 

c: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

d: \(x^3-7x-6\)

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)\)

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

Bài 8:

a: Xét tứ giác AEHF có

\(\widehat{AEH}=\widehat{AFH}=\widehat{FAE}=90^0\)

=>AEHF là hình chữ nhật

=>AH=FE

Xét ΔAMN vuông tại A có AH là đường cao

nên \(AH^2=HM\cdot HN\)

=>\(FE^2=HM\cdot HN\)

b: Ta có: AEHF là hình chữ nhật

=>\(\widehat{AFE}=\widehat{AHE}\)

mà \(\widehat{AHE}=\widehat{M}\left(=90^0-\widehat{HAM}\right)\)

nên \(\widehat{AFE}=\widehat{M}\)

Ta có; ΔAMN vuông tại A

mà AD là đường trung tuyến

nên DN=DA

=>\(\widehat{DAN}=\widehat{DNA}\)

Ta có: \(\widehat{AFE}+\widehat{DAN}\)

\(=\widehat{DNA}+\widehat{M}\)

\(=90^0\)

=>AD\(\perp\)FE