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1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)
2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)
3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)
4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)
5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)
c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)
e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
a) \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)
b) \(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}=\dfrac{\sqrt{3}+1-\left(\sqrt{3}-1\right)}{3-1}=1\)
c) \(2\sqrt{5}-3\sqrt{45}+\sqrt{500}=2\sqrt{5}-9\sqrt{5}+10\sqrt{5}=3\sqrt{5}\)
d) \(\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\sqrt{3}=\dfrac{1-\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1-3-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-2-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=-\sqrt{2}\)
e) \(\dfrac{1}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}+5\sqrt{3}=\dfrac{2-\sqrt{3}-\left(2+\sqrt{3}\right)}{4-3}+5\sqrt{3}=-2\sqrt{3}+5\sqrt{3}=3\sqrt{3}\)
f) \(\sqrt{3}-\sqrt{4+2\sqrt{3}}=\sqrt{3}-\left(\sqrt{3}+1\right)=-1\)
g) \(\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\sqrt{5}-\dfrac{4}{\sqrt{5}+1}=\dfrac{5+\sqrt{5}-4}{\sqrt{5}+1}=1\)
h)\(\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}=\sqrt{37-20\sqrt{3}+\left(5+2\sqrt{3}\right)}=\sqrt{42-18\sqrt{3}}=\sqrt{\left(3\sqrt{3}+3\right)^2+6}\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
a) \(\sqrt{20}+2\sqrt{45}-3\sqrt{80}+\sqrt{125}=\sqrt{4.5}+2\sqrt{9.5}-3\sqrt{16.5}+\sqrt{25.5}=2\sqrt{5}+6\sqrt{5}-12\sqrt{6}+5\sqrt{5}=\sqrt{5}\) b) \(\sqrt{6+2\sqrt{5}}-\sqrt{21+4\sqrt{5}}+\sqrt{5}\left(\sqrt{5}+1\right)=\sqrt{5+2\sqrt{5}+1}-\sqrt{20+2.2\sqrt{5}+1}+\sqrt{5}\left(\sqrt{5}+1\right)=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(2\sqrt{5}+1\right)^2}+\sqrt{5}\left(\sqrt{5}+1\right)\) = / \(\sqrt{5}+1\) / + / \(2\sqrt{5}+1\) / \(+\sqrt{5}\left(\sqrt{5}+1\right)\)
\(=\sqrt{5}+1\) + \(2\sqrt{5}+1\) \(+\sqrt{5}\left(\sqrt{5}+1\right)\)
= \(4\sqrt{5}+7\)
Bạn mới trả lời câu a,b thôi đúng không?Nhưng mà rất cảm ơn bạn!:)))
a: \(A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot A\cdot\sqrt[3]{4-5}\)
\(\Leftrightarrow A^3=4-3A\)
=>A=1
c: \(C=1+\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(=1+3=4\)
a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
Bài 1:
a: \(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
\(=5\sqrt{5}-4\sqrt{5}-12\sqrt{5}+12\sqrt{5}=\sqrt{5}\)
b: \(\sqrt{\left(1-2\sqrt{7}\right)^2}+\sqrt{8+2\sqrt{7}}\)
\(=2\sqrt{7}-1+\sqrt{7}+1=3\sqrt{7}\)
c:\(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\)
\(=\dfrac{1+\sqrt{3}-1+\sqrt{3}}{-2}=-\dfrac{2\sqrt{3}}{2}=-\sqrt{3}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{6+\sqrt{20}}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}\)
C = \(\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)}{10-2}\)
C = \(\dfrac{2\sqrt{30-10\sqrt{5}}+2\sqrt{6-2\sqrt{5}}}{8}\)
C = \(\dfrac{2\sqrt{\left(5-\sqrt{5}\right)^2}+2\sqrt{\left(\sqrt{5}-1\right)^2}}{8}\)
C = \(\dfrac{2\left(5-\sqrt{5}\right)+2\left(\sqrt{5}-1\right)}{8}\)
C = \(\dfrac{10-2\sqrt{5}+2\sqrt{5}-2}{8}\) = \(\dfrac{8}{8}\) = \(1\)
D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
D = \(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
D = \(7-3\sqrt{5}-\left(7+3\sqrt{5}\right)\) = \(7-3\sqrt{5}-7-3\sqrt{5}\)
D = \(-6\sqrt{5}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\) = \(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)
\(=2\sqrt{5}+3\sqrt{5}-3\sqrt{5}-\sqrt{5}=\sqrt{5}\)