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\(a,ĐK:x\ge0;x\ne9\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\\ b,x=13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\\ \Leftrightarrow A=\dfrac{-3}{2\sqrt{3}-1+3}=\dfrac{-3}{2\sqrt{3}+2}=\dfrac{-3\left(2\sqrt{3}-2\right)}{8}\)
\(c,A< -\dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\\ d,A=-\dfrac{2}{3}\Leftrightarrow\dfrac{3}{\sqrt{x}+3}=\dfrac{2}{3}\\ \Leftrightarrow2\sqrt{x}+6=9\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\\ e,\Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}=0\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x=0\left(tm\right)\\ f,\sqrt{x}+3\ge3\\ \Leftrightarrow A=-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{3}{3}=-1\\ A_{min}=-1\Leftrightarrow x=0\)
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a: \(A=\dfrac{x+2\sqrt{x}+x-3\sqrt{x}+2-x-\sqrt{x}-2}{x-4}\)
\(=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
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ĐK: \(x\ne\pm3\)
Khi đó:
\(C=\dfrac{2\left(x-3\right)}{x^2-9}+\dfrac{1\left(x+3\right)}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6}{x^2-9}+\dfrac{x+3}{x^2-9}-\dfrac{8}{x^2-9}\\ =\dfrac{2x-6+x+3-8}{x^2-9}\\ =\dfrac{3x-11}{x^2-9}\)
Thế x = 4 vào C được:
\(C=\dfrac{3.4-11}{4^2-9}=\dfrac{12-11}{16-9}=\dfrac{1}{7}\)
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\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)
\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)
\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)
\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)
B1.
Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)
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Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn!
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a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne4\\x\ge0\end{matrix}\right.\)
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\(T=\dfrac{2\left(x-1\right)}{\sqrt{x}+1}+\dfrac{x-4}{\sqrt{x}-2}\)
\(T=\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)
\(T=2\left(\sqrt{x}+1\right)+\left(\sqrt{x}+2\right)\)
\(T=2\sqrt{x}+2+\sqrt{x}+2\)
\(T=3\sqrt{x}+4\)
\(x=4\)
\(\Rightarrow T=3\sqrt{4}+4=3.2+4=10\)