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\(A=\dfrac{\sqrt{60}}{\sqrt{15}}=\sqrt{\dfrac{60}{15}}=\sqrt{4}=2\)
\(B=\sqrt{\dfrac{72}{15}}:\sqrt{\dfrac{2}{15}}=\sqrt{\dfrac{72}{15}}\cdot\sqrt{\dfrac{15}{2}}=\sqrt{\dfrac{72}{2}}=\sqrt{36}=6\)
\(C=\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{2}-\sqrt{3}\right)=\left(\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2=2-3=-1\)
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Ta có: \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
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Bài I:
1: Thay x=4 vào A, ta được:
\(A=\dfrac{4}{2+1}=\dfrac{4}{3}\)
2: \(B=\dfrac{3}{\sqrt{x}+1}+\dfrac{x+5}{x-1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3}{\sqrt{x}+1}+\dfrac{\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)+x+5-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\sqrt{x}-3+x-\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
3: P=A*B
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\)
P<=4
=>P-4<=0
=>\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-1}< =0\)
=>\(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}< =0\)
=>\(\sqrt{x}-1< 0\)
=>\(\sqrt{x}< 1\)
=>0<=x<1
Kết hợp ĐKXĐ, ta được: 0<=x<1
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\(a,\Leftrightarrow\left\{{}\begin{matrix}1-4m=-\dfrac{1}{2}\\m-2\ne3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=\dfrac{3}{8}\\m\ne5\end{matrix}\right.\Leftrightarrow m=\dfrac{3}{8}\\ b,\Leftrightarrow1-4m>0\Leftrightarrow m< \dfrac{1}{4}\\ c,\Leftrightarrow x=\dfrac{1}{2};y=0\Leftrightarrow\dfrac{1}{2}\left(1-4m\right)=2-m\Leftrightarrow1-4m=4-2m\\ \Leftrightarrow m=-\dfrac{3}{2}\)
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1,\(\sqrt{\left(x-1\right)^2}=\left|x-1\right|=-\left(x-1\right)=1-x\)
2,\(\sqrt{\left(a-2b\right)^2}=\left|a-2b\right|=-\left(a-2b\right)=2b-a\)
3,\(\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)
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