Thực hiện phép tính
\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}+1}\)
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\(\sqrt[3]{24}+5\cdot\frac{\sqrt[3]{81}}{\sqrt[3]{24}}-7\sqrt[3]{192}+\sqrt[3]{\frac{1}{9}}+\sqrt[3]{375}\) (Thực hiện phép tính )
\(=2\sqrt[3]{3}+5\cdot\dfrac{3}{2}-7\cdot4\sqrt[3]{3}+\dfrac{1}{3}\sqrt[3]{3}+5\sqrt[3]{3}\)
\(=-\dfrac{62}{3}\sqrt[3]{3}+\dfrac{15}{2}\)
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Bài 1. thực hiện phép tínha) sqrt{frac{5+sqrt{21}}{5-sqrt{21}}}+sqrt{frac{5-sqrt{21}}{5+sqrt{21}}} b) sqrt{frac{4+sqrt{7}}{4-sqrt{7}}}+sqrt{frac{4-sqrt{7}}{4+sqrt{7}}}Bài 2. Tính:a) Mleft(2sqrt{2}right)sqrt{2+4sqrt{3+sqrt{2}+sqrt{11-6sqrt{2}}}}b) frac{1}{sqrt{25}+sqrt{24}}+frac{1}{sqrt{24}+sqrt{23}}+...+frac{1}{sqrt{2}+sqrt{1}}4c)frac{left(5sqrt{3}+sqrt{50}right)left(5-sqrt{24}right)}{sqrt{75}0-5sqrt{2}}
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Bài 1. thực hiện phép tính
a) \(\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\) b) \(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
Bài 2. Tính:a) \(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
b) \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
c)
\(\frac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}0-5\sqrt{2}}\)
\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)
\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)
Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)
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\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)
\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)
\(=\sqrt{16+32\sqrt{6}}\)
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THỰC HIỆN PHÉP TÌNH VÀ RÚT GỌN CÁC BIỂU THỨC:
A=\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{50}{3}}-\sqrt{24}\right).\)\(\sqrt{6}\)
B= \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right)\)\(:\frac{1}{\sqrt{7}-\sqrt{5}}\)
Bài 1. (2,0 điểm) Thực hiện phép tính: n) 7/9 * sqrt(81) - 1/2 * sqrt(16) . c) (sqrt(8/3) - sqrt(24) + sqrt(50/3)) , sqrt 12 . » sqrt((sqrt(7) - 4) ^ 2) + sqrt(7) 1/(5 + 2sqrt(3)) + 1/(5 - 2sqrt(3))
Thực hiện các phép tính sau
a, \(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
b, \(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
c, \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
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a, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)
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b. = \(\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}-\frac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}-\frac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}\)
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Bài 7 Thực hiện phép tính
a)\(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\))
d)\(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
...
thực hiện phép tính:
1, sqrt{5+2sqrt{24}}-sqrt{2}
2, frac{3-2sqrt{3}}{sqrt{3}-2} 3, frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}
4, frac{1}{1-sqrt{2}}-frac{1}{1+sqrt{2}} 5, frac{sqrt{12}-6}{sqrt{8}-sqrt{24}}-frac{3-sqrt{3}}{sqrt{3}}-frac{4}{1-sqrt{7}}
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thực hiện phép tính:
1, \(\sqrt{5+2\sqrt{24}}-\sqrt{2}\)
2, \(\frac{3-2\sqrt{3}}{\sqrt{3}-2}\) 3, \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\)
4, \(\frac{1}{1-\sqrt{2}}-\frac{1}{1+\sqrt{2}}\) 5, \(\frac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\frac{3-\sqrt{3}}{\sqrt{3}}-\frac{4}{1-\sqrt{7}}\)
a/ \(\sqrt{5+\sqrt{24}}-\sqrt{2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{2}=\left|\sqrt{3}+\sqrt{2}\right|-\sqrt{2}=\sqrt{3}+\sqrt{2}-\sqrt{2}=\sqrt{3}\)
b/ \(\frac{3-2\sqrt{3}}{\sqrt{3}-2}=\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}=\sqrt{3}\)
c/ \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
d/ \(\frac{1}{1-\sqrt{2}}-\frac{1}{1+\sqrt{2}}=\frac{1+\sqrt{2}-1+\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\frac{2\sqrt{2}}{1-2}=-2\sqrt{2}\)
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Thực hiện phép tính: \(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
\(\left(\sqrt{\frac{1}{7}}+\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{\sqrt{7}}{7}+\frac{4\sqrt{7}}{7}+\sqrt{7}\right):\sqrt{7}\)
= \(\left(\frac{5\sqrt{7}}{7}+\sqrt{7}\right):\sqrt{7}=\frac{5\sqrt{7}}{7}.\frac{1}{\sqrt{7}}+1=\frac{5}{7}+1=\frac{12}{7}\)
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Tính:
\(a)\frac{1}{\sqrt{7-\sqrt{24}}+1}+\frac{1}{\sqrt{7+\sqrt{24}}-1}\\ b)\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+...+\frac{1}{\sqrt{2014}+\sqrt{2015}}\)
\(a=\frac{1}{\sqrt{7-2\sqrt{6}}+1}+\frac{1}{\sqrt{7+2\sqrt{6}}-1}=\frac{1}{\sqrt{\left(\sqrt{6}-1\right)^2}+1}+\frac{1}{\sqrt{\left(\sqrt{6}+1\right)^2}-1}\)
\(=\frac{1}{\sqrt{6}}+\frac{1}{\sqrt{6}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}\)
Coi lại đề câu b, quy luật ở số hạng cuối cùng sai (nhìn 2 số hạng đầu 2 số dưới căn hơn kém nhau 4 đơn vị, số cuối lại chỉ hơn kém nhau 1 đơn vị)
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