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a: AB và AC là hai tiếp tuyến cắt nhau tại A
b: Xét tứ giác OBAC có
\(\widehat{OBA}+\widehat{OCA}=90^0+90^0=180^0\)
=>OBAC là tứ giác nội tiếp
=>O,B,A,C cùng thuộc 1 đường tròn
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5.
a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=5\left(cm\right)\)
Áp dụng HTL: \(AB^2=BH\cdot BC\Rightarrow BH=\dfrac{AB^2}{BC}=1,8\left(cm\right)\)
b, \(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{4}{5}\approx\sin53^0\Leftrightarrow\widehat{B}\approx53^0\)
Vì tg ABC vuông tại A nên \(\widehat{C}=90^0-\widehat{B}=37^0\)
c, Áp dụng HTL: \(AH\cdot BC=AB\cdot AC\Rightarrow AH=\dfrac{AB\cdot AC}{BC}\)
\(\Rightarrow AH\cdot AC=\dfrac{AB\cdot AC^2}{BC}=\dfrac{AB\cdot CH\cdot BC}{BC}=AB\cdot CH\)
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Bài 4:
ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{x^2-9}-\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
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a) \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) \(P=\dfrac{1}{2}\Rightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\Rightarrow4-10\sqrt{x}=\sqrt{x}+3\Rightarrow11\sqrt{x}=1\)
\(\Rightarrow x=\dfrac{1}{121}\)
c) \(P\le\dfrac{2}{3}\Rightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\le\dfrac{2}{3}\Rightarrow\dfrac{2}{3}-\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\ge0\)
\(\Rightarrow\dfrac{2\left(\sqrt{x}+3\right)-3\left(2-5\sqrt{x}\right)}{3\left(\sqrt{x}+3\right)}\ge0\Rightarrow\dfrac{17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\ge0\) (luôn đúng)
Bài 1:
a) Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) Để \(P=\dfrac{1}{2}\) thì \(4-10\sqrt{x}-\sqrt{x}-3=0\)
\(\Leftrightarrow-11\sqrt{x}=-1\)
\(\Leftrightarrow x=\dfrac{1}{121}\)
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Câu 2: b. \(\sqrt{9x^2-6x+1}=9\)
<=> \(\sqrt{\left(3x-1\right)^2}=9\)
<=> 3x - 1 = 9
<=> 3x = 10
<=> x = \(\dfrac{10}{3}\)