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Bài 7:
Ta có: \(P=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)
\(=\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}}{x\sqrt{x}-1}\)
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a, \(2\sqrt{3}-\sqrt{4+x^2}=0\Leftrightarrow\sqrt{4+x^2}=2\sqrt{3}\)
\(\Leftrightarrow x^2+4=12\Leftrightarrow x^2=8\Leftrightarrow x=\pm2\sqrt{2}\)
b, \(\sqrt{16x+16}-\sqrt{9x+9}=0\)ĐK : x >= -1
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=0\Leftrightarrow\sqrt{x+1}=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
c, \(\sqrt{4\left(x+2\right)^2}=8\Leftrightarrow2\left|x+2\right|=8\Leftrightarrow\left|x+2\right|=4\)
TH1 : \(x+2=4\Leftrightarrow x=2\)
TH2 : \(x+2=-4\Leftrightarrow x=-6\)
c: Ta có: \(\sqrt{4\left(x+2\right)^2}=8\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
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b: \(=\dfrac{x\left(\sqrt{x-1}-1+\sqrt{x-1}+1\right)}{1+x-2}=\dfrac{x\cdot2\sqrt{x-1}}{x-1}=\dfrac{2x}{\sqrt{x-1}}\)
c:
Sửa đề: 1/căn x
\(=\dfrac{3\sqrt{x}-x+x+9}{9-x}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\sqrt{x}}{\sqrt{x}+4}\)
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a: Xét (O) có
ΔAMB nội tiếp đường tròn
AB là đường kính
Do đó: ΔAMB vuông tại M
Xét tứ giác AMCK có
\(\widehat{AKC}+\widehat{AMC}=180^0\)
nên AMCK là tứ giác nội tiếp
hay A,M,C,K cùng thuộc một đường tròn
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a) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=x-\sqrt{x}+1\)
b) \(P=x-\sqrt{x}+1=\left(\sqrt{x}\right)^2-2.\sqrt{x}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow P_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)
c) \(Q=\dfrac{2\sqrt{x}}{P}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\)
Ta có: \(\left\{{}\begin{matrix}2\sqrt{x}>0\left(x>0\right)\\x+\sqrt{x}+1>0\end{matrix}\right.\Rightarrow Q>0\)
Lại có: \(3x-5\sqrt{x}+3=3\left(\left(\sqrt{x}\right)^2-2.\sqrt{x}.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2\right)+\dfrac{11}{12}\)
\(=3\left(\sqrt{x}-\dfrac{5}{6}\right)^2+\dfrac{11}{12}>0\)
\(\Rightarrow3x-5\sqrt{x}+3>0\Rightarrow3x-3\sqrt{x}+3>2\sqrt{x}\Rightarrow3\left(x-\sqrt{x}+1\right)>2\sqrt{x}\)
\(\Rightarrow3>\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}\Rightarrow Q< 3\Rightarrow0< Q< 3\)
mà \(Q\in Z\Rightarrow Q\in\left\{1;2\right\}\)
Từ\(Q\) tính ta x thôi
a, \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)ĐK : \(x>0;x\ne1\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2\)
\(=x-\sqrt{x}\)
b, Ta có : \(x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu ''='' xảy ra khi \(x=\dfrac{1}{4}\)
Vậy GTNN P là -1/4 khi x = 1/4
c, Ta có : \(G=\dfrac{2\sqrt{x}}{P}\Rightarrow G=\dfrac{2\sqrt{x}}{x-\sqrt{x}}=\dfrac{2}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\sqrt{x}-1\) | 1 | -1 | 2 | -2 |
\(\sqrt{x}\) | 2 | 0 | 3 | -1 |
x | 4 | 0 ( loại ) | 9 | loại |
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\(AB=\tan C\cdot AC=\tan40^0\cdot100\approx84\left(m\right)\)
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