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\(\Leftrightarrow x=0\Leftrightarrow\left\{{}\begin{matrix}5-m=y\\3+m=y\end{matrix}\right.\Leftrightarrow y=4\Leftrightarrow5-m=4\Leftrightarrow m=1\)
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7:
a: ĐKXĐ: x>=0; x<>1
\(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\)
b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)
c: |D|=1/3
=>D=-1/3 hoặc D=1/3
=>\(\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{3}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)
=>\(\sqrt{x}+1=3\)
=>\(\sqrt{x}=2\)
=>x=4
6:
a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{3+\sqrt{x}}\cdot\dfrac{-\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
b: C<-1
=>C+1<0
=>\(\dfrac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)
=>\(-\sqrt{x}+4< 0\)
=>\(-\sqrt{x}< -4\)
=>\(\sqrt{x}>4\)
=>x>16
\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Để `C < -1` Ta có :
\(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)
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Bài 5:
\(P=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\) (đk:\(a>0;a\ne2;a\ne1\))
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right).\dfrac{a-2}{a+2}\)
\(=\left(\sqrt{a}+1+\dfrac{1}{\sqrt{a}}-\sqrt{a}+1-\dfrac{1}{\sqrt{a}}\right).\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)
b) \(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
Để \(P\in Z\) \(\Leftrightarrow\dfrac{8}{a+2}\in Z\)
Có \(a\in Z,a>0\) \(\Rightarrow a+2\in Z\) và \(a+2>2\)
=> \(a+2\inƯ\left(8\right)=\left\{4;8\right\}\) \(\Leftrightarrow a\in\left\{2;6\right\}\) mà a\(\ne2\) =>a=6
Vậy a=6
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a: \(\sqrt{3}-2=\sqrt{3}-\sqrt{4}< 0\)
=>Hàm số \(y=\left(\sqrt{3}-2\right)x+3\) nghịch biến trên R
a: Khi \(x=2+\sqrt{3}\) thì
\(y=\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)+3\)
=3-4+3
=6-4
=2
b: Thay \(y=\sqrt{3}+1\) vào \(y=\left(\sqrt{3}-2\right)x+3\), ta được:
\(x\left(\sqrt{3}-2\right)+3=\sqrt{3}+1\)
=>\(x\left(\sqrt{3}-2\right)=\sqrt{3}-2\)
=>x=1
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Bài 4:
a: Xét tứ giác OBAC có
\(\widehat{OBA}+\widehat{OCA}=180^0\)
Do đó: OBAC là tứ giác nội tiếp
b: Xét (O) có
AB là tiếp tuyến
AC là tiếp tuyến
Do đó: AB=AC
hay A nằm trên đường trung trực của BC(1)
Ta có: OB=OC
nên O nằm trên đường trung trực của BC(2)
Từ (1) và (2) suy ra OA là đường trung trực của BC
hay OA⊥BC
c: Xét ΔOBA vuông tại B có BA là đường cao
nên \(OH\cdot OA=OB^2=R^2\)
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ĐKXĐ: \(x>0;x\ne1\)
\(P=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\left(\sqrt{x}-1\right)=\dfrac{x-1}{\sqrt{x}}\)
b.
\(P>0\Leftrightarrow\dfrac{x-1}{\sqrt{x}}>0\)
\(\Rightarrow x-1>0\Rightarrow x>1\)