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\(A=\sqrt{4^2b}+2\sqrt{2^2\cdot10b}-3\sqrt{3^2\cdot10b}=4\sqrt{b}+4\sqrt{10}\cdot\sqrt{b}-9\sqrt{10}\cdot\sqrt{b}\)
\(=4\sqrt{b}-5\sqrt{10}\sqrt{b}=\left(4-5\sqrt{10}\right)\sqrt{b}\)
Rut gon A = √16b+2√40b−3√90bva`b≥0
A=√42b+2√22·10b−3√32·10b=4√b+4√10·√b−9√10·√b
=4√b−5√10√b=(4−5√10)√b
a) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\) = \(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\) = \(-\sqrt{3}\)
b) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\) = \(7\sqrt{2}-6\sqrt{2}+\sqrt{2}\) = \(2\sqrt{2}\)
c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\) = \(3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\) = \(6\sqrt{a}\)
d) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\) = \(4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
= \(4\sqrt{b}-5\sqrt{10b}\)
Bài 1:
\(\sqrt{27a^2}=3a\sqrt{3}\)
Bài 2:
\(\dfrac{2}{3}\sqrt{3xy}=\sqrt{3xy\cdot\dfrac{4}{9}}=\sqrt{\dfrac{4}{3}xy}\)
Bài 3:
\(=4\sqrt{b}+2\cdot2\sqrt{10b}-3\cdot3\sqrt{10b}=4\sqrt{b}-5\sqrt{10b}\)
\(=7\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}=7\sqrt{b}-5\sqrt{10b}\)
1.
\(x-6\sqrt{x}-\sqrt{x}+6=0\)
\(\Leftrightarrow\left(\sqrt{x}-6\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=36\\x=1\end{cases}}\)
2.
\(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\\sqrt{x-3}=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}}\)
c,C= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\left(x\ge1\right)\)
=\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
=\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\) (1)
TH1: \(\sqrt{x-1}< 1\) hay \(1\le x< 2\)
Từ (1)=>C= \(\sqrt{x-1}+1+1-\sqrt{x-1}\)=2
TH2: \(\sqrt{x-1}\ge1\) hay \(x\ge2\)
Từ (1) =>C=\(\sqrt{x-1}+1+\sqrt{x-1}-1\)=\(2\sqrt{x-1}\)
d, D=\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{13+30\sqrt{2}+\sqrt{8+2\sqrt{8}+1}}=\sqrt{13+30\sqrt{2}+\sqrt{\left(\sqrt{8}+1\right)^2}}\)
=\(\sqrt{13+30\sqrt{2}+\sqrt{8}+1}=\sqrt{14+30\sqrt{2}+2\sqrt{2}}\)
=\(\sqrt{14+32\sqrt{2}}\)
a)\(\frac{x-y}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
b)\(\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
\(\sqrt{16b}+2\sqrt{40b}-30\sqrt{90b}\)
\(=\sqrt{16}\sqrt{b}+2\sqrt{40}\sqrt{b}-30\sqrt{90}\sqrt{b}\)
\(=\sqrt{b}\left(\sqrt{16}+2\sqrt{40}-30\sqrt{90}\right)\)
\(=\sqrt{b}\left(4+4\sqrt{10}-90\sqrt{10}\right)\)
\(=\sqrt{b}\left(4-86\sqrt{10}\right)\)