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a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
a) = \(5\sqrt{2}-3\sqrt{6}+3\sqrt{2}+5\sqrt{6}\)
= \(8\sqrt{2}+2\sqrt{6}\)
b) = \(2\sqrt{3}-4\sqrt{2}-5\sqrt{3}-\sqrt{2}\)
= \(-3\sqrt{3}-5\sqrt{2}\)
c) = \(\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=\(\frac{2\sqrt{2}+2-2-\sqrt{2}}{2^2-\sqrt{2^2}}\)
=\(\frac{\sqrt{2}}{4-2}\) = \(\frac{\sqrt{2}}{2}\)
d) = \(2\sqrt{6}-5\sqrt{6}+2\sqrt{2}\)
=\(-3\sqrt{6}+2\sqrt{2}\)
e) = \(8\sqrt{6}+3\sqrt{6}-6\sqrt{6}=5\sqrt{6}\)
f) = \(4\sqrt{3}+9\sqrt{3}-4\sqrt{3}=9\sqrt{3}\)
g) = \(10+5\sqrt{10}-5\sqrt{10}=10\)
h) = \(\frac{\left(3+\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{\left(3-\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
= \(\frac{9+3\sqrt{3}+3\sqrt{3}+3}{3^2-\sqrt{3^2}}+\frac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-\sqrt{3^2}}\)
= \(\frac{12+6\sqrt{3}}{9-3}+\frac{12-6\sqrt{3}}{9-3}\)
= \(\frac{12+6\sqrt{3}+12-6\sqrt{3}}{6}\)
= \(\frac{24}{6}=4\)
k) = \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
l) = \(\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\)
= \(\frac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{\sqrt{8^2}-2^2}\)
= \(\frac{2\sqrt{6}}{8-4}=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\)
Bài 1:
a, \(4\sqrt{3+2\sqrt{2}}-\sqrt{57+40\sqrt{2}}\)
\(=4\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(4\sqrt{2}+5\right)^2}\)
\(=4\left(\sqrt{2}+1\right)-4\sqrt{2}-5\)
\(=4\sqrt{2}+4-4\sqrt{2}-5=-1\)
b, \(B=\sqrt{1100}-7\sqrt{44}+2\sqrt{176}-\sqrt{1331}\)
\(=10\sqrt{11}-14\sqrt{11}+8\sqrt{11}-11\sqrt{11}=-7\sqrt{11}\)
c, \(C=\sqrt{\left(1-\sqrt{2002}\right)^2}.\sqrt{2003+2\sqrt{2002}}\)
\(=\left(1-\sqrt{2002}\right).\sqrt{\left(\sqrt{2002}+1\right)^2}\)
\(=\left(1-\sqrt{2002}\right).\left(\sqrt{2002}+1\right)=-2001\)
Câu d bạn kiểm tra lại đề bài nhé.
Bài 2:
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
a, ĐK: \(x\ge0,x\ne1\)
b, ĐK: \(x\ge0,x\ne1\)
\(A=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}+\frac{\sqrt{x}}{1-x}\)
\(=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{2}+2}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{2\sqrt{x}+2-2\sqrt{x}+2}{4\left(x-1\right)}-\frac{\sqrt{x}}{x-1}\)
\(=\frac{4-4\sqrt{x}}{4\left(x-1\right)}=\frac{4\left(1-\sqrt{x}\right)}{4\left(1-x\right)}=\frac{1-\sqrt{x}}{1-x}\)
Thay \(x=3\left(TM\right)\)vào A ta có: \(A=\frac{1-\sqrt{3}}{3-1}=\frac{1-\sqrt{3}}{2}\)
Vậy với \(x=3\)thì \(A=\frac{1-\sqrt{3}}{2}\)
c, \(\left|A\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}A=\frac{1}{2}\\A=-\frac{1}{2}\end{cases}}\)
TH1: \(A=\frac{1}{2}\)\(\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=\frac{1}{2}\Leftrightarrow2-2\sqrt{x}=x-1\)\(\Leftrightarrow x-1-2+2\sqrt{x}=0\)\(\Leftrightarrow x+2\sqrt{x}-3=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\\sqrt{x}=-3\left(L\right)\end{cases}}}\)
TH2: \(A=-\frac{1}{2}\Leftrightarrow\frac{1-\sqrt{x}}{x-1}=-\frac{1}{2}\)\(\Leftrightarrow2-2\sqrt{x}=1-x\Leftrightarrow-x+1-2+2\sqrt{x}=0\)\(\Leftrightarrow-x-1+2\sqrt{x}=0\Leftrightarrow x-2\sqrt{x}+1=0\)\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\Leftrightarrow\sqrt{x}=-1\left(L\right)\)
Vậy với \(x=1\)thì \(\left|A\right|=\frac{1}{2}\)
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
a) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)
\(=\frac{3}{2}\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)
\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)
b) \(=\frac{\sqrt{5}+3-\sqrt{5}+3}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}.\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\)
\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)
c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)
\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)
\(\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}-\frac{1+\sqrt{2}}{1-\sqrt{2}}\right):\sqrt{72}\)
\(\left[\frac{\left(1-\sqrt{2}\right)^2-\left(1+\sqrt{2}\right)^2}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\right]:\sqrt{72}\)
\(=\frac{1-2\sqrt{2}+2-1-2\sqrt{2}-2}{1-2}\cdot\frac{1}{\sqrt{72}}\)
\(=\frac{-2\sqrt{2}-2\sqrt{2}}{-1}\cdot\frac{1}{\sqrt{72}}\)
\(=4\sqrt{2}\cdot\frac{1}{2\sqrt{18}}=\frac{2}{\sqrt{9}}=\frac{2}{3}\)