Cần gấp bài 2,3,4
Bài 2:
1: Thay x=169 vào A, ta được:
\(A=\dfrac{2\cdot13+5}{13-1}=\dfrac{26+5}{12}=\dfrac{31}{12}\)
2: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+2\left(\sqrt{x}-2\right)-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+4\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
3: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+5}{\sqrt{x}-1}=\dfrac{2\sqrt{x}+5}{\sqrt{x}+3}\)
\(P-1=\dfrac{2\sqrt{x}+5}{\sqrt{x}+3}-1=\dfrac{2\sqrt{x}+5-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}>0\)
=>P>1
=>\(\sqrt{P}-1>0\)
=>\(\sqrt{P}\left(\sqrt{P}-1\right)>0\)
=>\(P>\sqrt{P}\)
Bài 3:
1: Khi m=-1 thì \(y=\left(-1-2\right)x+\left(-1\right)-1=-3x-2\)
Vẽ đồ thị:
2: Tọa độ điểm A là:
\(\left\{{}\begin{matrix}x=0\\y=\left(m-2\right)\cdot0+m-1=m-1\end{matrix}\right.\)
=>A(0;m-1)
\(OA=\sqrt{\left(0-0\right)^2+\left(m-1-0\right)^2}=\left|m-1\right|\)
Tọa độ điểm B là:
\(\left\{{}\begin{matrix}y=0\\x\left(m-2\right)+m-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=0\\x\left(m-2\right)=-m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{-m+1}{m-2}\end{matrix}\right.\)
\(OB=\sqrt{\left(\dfrac{-m+1}{m-2}-0\right)^2+\left(0-0\right)^2}=\sqrt{\left(\dfrac{-m+1}{m-2}\right)^2}=\dfrac{\left|m-1\right|}{\left|m-2\right|}\)
ΔOAB vuông tại O
=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot\left|m-1\right|\cdot\dfrac{\left|m-1\right|}{\left|m-2\right|}=\dfrac{\dfrac{1}{2}\left(m-1\right)^2}{\left|m-2\right|}\)
Để \(S_{OAB}=1\) thì \(\dfrac{1}{2}\cdot\dfrac{\left(m-1\right)^2}{\left|m-2\right|}=1\)
=>\(\left(m-1\right)^2=2\left|m-2\right|\)(1)
TH1: m>=2
(1) sẽ trở thành: \(\left(m-1\right)^2=2\left(m-2\right)\)
=>\(m^2-2m+1-2m+4=0\)
=>\(m^2-4m+4+1=0\)
=>\(\left(m-2\right)^2+1=0\)(vô lý)
=>Loại
TH2: m<2
(1) sẽ trở thành \(\left(m-1\right)^2=-2\left(m-2\right)\)
=>\(m^2-2m+1+2m-4=0\)
=>\(m^2=3\)
=>\(\left[{}\begin{matrix}m=\sqrt{3}\left(nhận\right)\\m=-\sqrt{3}\left(nhận\right)\end{matrix}\right.\)