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\(P=\sqrt{\frac{15}{2}}.\sqrt{\frac{10.\left(a-1\right)^2}{3}}\) ( ĐK a<1 )
\(\Leftrightarrow P=\frac{\sqrt{15}}{\sqrt{2}}.\frac{\sqrt{10}.\sqrt{\left(a-1\right)^2}}{\sqrt{3}}\)
\(\Leftrightarrow P=\frac{\sqrt{15}.\sqrt{2}}{2}.\frac{\sqrt{10}.\sqrt{3}.\left|a-1\right|}{3}\)
\(\Leftrightarrow P=\frac{\sqrt{30}}{2}.\frac{\sqrt{30}\left(1-a\right)}{3}\)( vì a-1<0)
\(\Leftrightarrow P=\frac{\sqrt{30}.\sqrt{30}\left(1-a\right)}{2.3}\)
\(\Leftrightarrow\frac{30\left(1-a\right)}{6}\)
\(\Leftrightarrow5\left(1-a\right)\)
a: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}+3\right)+2\sqrt{2}-3}{-1}\)
\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)
b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)
\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)
c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)
\(=\dfrac{-2\left(3-2\sqrt{2}\right)+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{-6+4\sqrt{2}+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}\)
\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
a) ĐKXĐ: \(x\ne4\)và \(x>0\)
............................
\(\Leftrightarrow A=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}+2}\right)\)\(:\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{3x-6\sqrt{x}\left(\sqrt{x}+2\right)+3\sqrt{x}\left(\sqrt{x}-2\right)}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}:\left(\frac{x-2+10-x}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{3x-6x-12\sqrt{x}+3x-6\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\left(\frac{8}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{-18\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}-2}{8}\)
\(\Leftrightarrow A=\frac{-3}{4\left(\sqrt{x}+2\right)}\)
Vậy \(A=\frac{-3}{4\left(\sqrt{x}-2\right)}\)với \(x>0\)và \(x\ne4\)
b)Ta có \(A< 2\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}< 2\)
\(\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}-2< 0\)
\(\Leftrightarrow\frac{-3-8\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-3-8\sqrt{x}-16}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-18-8\sqrt{x}}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow-18-8\sqrt{x}< 0\)( Vì \(4\left(\sqrt{x}-2\right)>0\)với \(\forall x\))
\(\Leftrightarrow\sqrt{x}< \frac{-9}{4}\)(Vô Nghiệm)
Vậy không có gtr nào của x thỏa mãn A<2