1) Rút gọn
\(A=\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\)
2) So sánh: \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\)và \(\sqrt{3}\)
Những câu hỏi liên quan
B1: rút gọn biểu thức:
a, \(\frac{\left(4\sqrt{21}-4\sqrt{15}-\sqrt{4}+\sqrt{10}\right)}{4\sqrt{6}-2+4\sqrt{15}-\sqrt{10}}\)
b, \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
c, \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
d, \(\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2+\sqrt{3}}\)
Rút gọn các biểu thức sau:
a) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\) b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
c) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right):\sqrt{6}\) d) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
Bài 1: Rút gọn biểu thức1) sqrt{12}-sqrt{27}+sqrt{48} 2) left(sqrt{25}+sqrt{20}-sqrt{80}right):sqrt{5}3) 2sqrt{27}-sqrt{frac{16}{3}}-sqrt{48}-sqrt{8frac{1}{3}} 4) frac{1}{sqrt{5}-sqrt{3}}-frac{1}{sqrt{5}+sqrt{3}}5) left(sqrt{125}-sqrt{12}-2sqrt{5}right)left(3sqrt{5}-sqrt{3}+sqrt{27}right) 6) left(3sqrt{20}-sqrt{125}-15sqrt{frac{1}{5}}right).sqrt{5}7) left(6sqrt{128}-frac{3}{5}sqrt{50}+7sqrt{8}right):3sqrt{2} 8) left(2sqrt{48}-frac{3}{2}sqrt{frac{4}{3}}+sqrt{27}right).2sqrt{3...
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Bài 1: Rút gọn biểu thức
1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\) 2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\) 4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) 6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)
7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\) 8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\) 10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\) 12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
13) \(\sqrt{15-6\sqrt{6}}\) 14) \(\sqrt{8-2\sqrt{15}}\) 15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)
Rút gọn bt
\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)
\(=\sqrt{\sqrt{5}^2+2\cdot\sqrt{5}\left(\sqrt{2}+1\right)+\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}\)
\(=\sqrt{5}+\sqrt{2}+1\)
\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)
Rút gọn\(B=\frac{3\sqrt{8}+2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\\ C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Rút gọn biểu thức :
\(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{2-\sqrt{3}}}.\sqrt{2+\sqrt{3}}\)
\(\left[\left(a-b\right)\sqrt{\frac{a+b}{a-b}}+a-b\right]\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)với a>b>0
Chứng minh rằng :
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
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tính \(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
a) rút gọn A
b) tính A với \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\)\(\left[\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right]\left[\frac{a+1}{\sqrt{a}}\right]\)
\(A=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\) \(\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\left(a+1\right)}{a-1}\)
ta có \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(a=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(a=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)
\(a=2\left(16-15\right)\)
\(a=2\)
khi đó \(A=\frac{4.2.\left(2+1\right)}{2-1}=8.3=24\)
vậy.....
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Rút gọn \(A=\left(\sqrt{6+\sqrt{20}}-2\sqrt{3-\sqrt{5}}+\sqrt{15-10\sqrt{2}}\right):\left(2+\sqrt{8}\right)\)
\(A=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{2}.\sqrt{6-2\sqrt{5}}+\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{2}\left(\sqrt{5}-1\right)+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{10}+\sqrt{2}+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}=\dfrac{1}{2}\)
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Rút gọn biểu thức 1) frac{sqrt{5+2sqrt{6}}+sqrt{8+2sqrt{15}}}{sqrt{7+2sqrt{10}}}2) left(2+frac{3+sqrt{3}}{sqrt{3}+1}right)left(2+frac{3-sqrt{3}}{sqrt{3}-1}right):left(sqrt{5}-2right)3) left(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)4) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+...+frac{1}{sqrt{99}+sqrt{100}}5) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...-frac{1}{sqrt{98}-sqrt{99}}+frac{1}{sqrt{99}-s...
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Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
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Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
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Sao làm hổng ai bảo đú.n/g vậy :(((
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