Rút gọn
\(\sqrt{8-2\cdot\sqrt{15}}\) + \(\sqrt{8+2\cdot\sqrt{15}}\)
Những câu hỏi liên quan
Rút gọn \(A=\sqrt{8+2\cdot\sqrt{10+2\cdot\sqrt{5}}}+\sqrt{8-2\cdot\sqrt{10-2\cdot\sqrt{5}}}\)
Tính
a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)
b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
a) \(\sqrt{6-\sqrt{11}}\cdot\sqrt{6+\sqrt{11}}\)
\(=\sqrt{\left(6-\sqrt{11}\right)\left(6+\sqrt{11}\right)}\)
\(=\sqrt{6^2-\left(\sqrt{11}\right)^2}\)
\(=\sqrt{36-11}\)
\(=\sqrt{25}\)
\(=\sqrt{5^2}\)
\(=5\)
b) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
\(=\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}\)
\(=\sqrt{8^2-\left(\sqrt{15}\right)^2}\)
\(=\sqrt{64-15}\)
\(=\sqrt{49}\)
\(=\sqrt{7^2}\)
\(=7\)
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a: \(=\sqrt{6^2-11}=\sqrt{25}=5\)
b: \(=\sqrt{8^2-15}=\sqrt{49}=7\)
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Rút gọn \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
bạn có thể phân tích 4 + căn 5 = căn ( 4 + căn 5) . căn ( 4 + căn 5)
và căn 10 - căn 6 = căn 2( căn 5 - căn 3)
Khi đó Biểu thức rút gọn trở thành
căn(4 - căn 15).căn(4+15) . căn (4 + căn 15) . căn 2(căn 5 - căn 3)
= căn (16 - 15) . căn (8 + 2.căn 15).(căn 5 - căn 3) = căn (3 + 2.căn 3. căn 5 + 5). (căn 5 - căn 3)
= căn [ (căn 3 + căn 5)^2 ] . (căn 5 - căn 3) = (căn 5 + căn 3)(căn 5 - căn 3) = 5 - 3 = 2
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TínhAleft(4+sqrt{15}right)left(sqrt{10}-sqrt{6}right)cdotsqrt{4-sqrt{15}}Bleft(3-sqrt{5}right)cdotsqrt{3+sqrt{5}}+left(3+sqrt{5}right)cdotsqrt{3-sqrt{5}}Csqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{ }}3}}Dsqrt{4+sqrt{15}}+sqrt{4-sqrt{15}}-2sqrt{3-sqrt{5}}Efrac{sqrt{15-10sqrt{2}}+sqrt{13+4sqrt{5}}-sqrt{11+2sqrt{10}}}{2sqrt{3+2sqrt{2}}+sqrt{9-4sqrt{2}}+sqrt{12+8sqrt{2}}}
Đọc tiếp
Tính
A=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
B=\(\left(3-\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\cdot\sqrt{3-\sqrt{5}}\)
C=\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{ }}3}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
E=\(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{5}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
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Câu 1: Thực hiện phép tính
a,left(sqrt{12}+3sqrt{15}-4sqrt{135}right)cdotsqrt{3}
b,sqrt{252}-sqrt{700}+sqrt{1008}-sqrt{448}
c,2sqrt{40sqrt{12}}-2sqrt{sqrt{75}}-3sqrt{5sqrt{48}}
Câu 2: Rút gọn
a,frac{9sqrt{5}+3sqrt{27}}{sqrt{5}+sqrt{3}}
b,frac{3sqrt{8}+2sqrt{12}+sqrt{20}}{3sqrt{18}-2sqrt{27}+sqrt{45}}
c,left(4+sqrt{15}right)cdotleft(sqrt{10}-sqrt{6}right)cdotsqrt{4-sqrt{15}}
Câu 3:So sánh
a,3+sqrt{5}và2sqrt{2}+sqrt{6}
b,2sqrt{3}+4và3sqrt{2}+sqrt{10}
c,18vàsqrt{15}cdotsqrt{17}
Đọc tiếp
Câu 1: Thực hiện phép tính
\(a,\left(\sqrt{12}+3\sqrt{15}-4\sqrt{135}\right)\cdot\sqrt{3}\\ b,\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\\ c,2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
Câu 2: Rút gọn
\(a,\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\\ b,\frac{3\sqrt{8}+2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\\ c,\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
Câu 3:So sánh
\(a,3+\sqrt{5}và2\sqrt{2}+\sqrt{6}\\ b,2\sqrt{3}+4và3\sqrt{2}+\sqrt{10}\\ c,18và\sqrt{15}\cdot\sqrt{17}\)
giúp mk mới mọi người ơi
câu 1 rút gọn
a .\(2.\sqrt{48}+\sqrt{27}+\sqrt{3}\)
b\(\sqrt{45.a^2}+\sqrt{8.b^2}+5\cdot\sqrt{5\cdot a\cdot a^2}-3\cdot b\cdot\sqrt{2\cdot b}\)
a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)
=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)
12\(\sqrt{3}\)
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TÍNH :Asqrt{3+sqrt{5+2sqrt{3}}}cdotsqrt{3-sqrt{5+2sqrt{3}}}Bsqrt{4+sqrt{8}}cdotsqrt{2+sqrt{2+sqrt{2}}}cdotsqrt{2-sqrt{2+sqrt{2}}}Csqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{3}}}}Dleft[4+sqrt{15}right]left[sqrt{10}-sqrt{6}right]cdotsqrt{4-sqrt{15}}Eleft[3-sqrt{5}right]cdotsqrt{3+sqrt{5}}text{ }+left[3+sqrt{5}right]cdotsqrt{3-sqrt{5}}
Đọc tiếp
TÍNH :
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}}\cdot\sqrt{3-\sqrt{5+2\sqrt{3}}}\)
\(B=\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(C=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(D=\left[4+\sqrt{15}\right]\left[\sqrt{10}-\sqrt{6}\right]\cdot\sqrt{4-\sqrt{15}}\)
\(E=\left[3-\sqrt{5}\right]\cdot\sqrt{3+\sqrt{5}}\text{ }+\left[3+\sqrt{5}\right]\cdot\sqrt{3-\sqrt{5}}\)
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}}=\sqrt{\left(3^2\right)-\left(\sqrt{5+2\sqrt{3}}\right)^2}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=2\)
\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}=\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
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\(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4^2-15}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(E=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
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cho số thực a<1. Rút gọn biểu thức \(P=\sqrt{\frac{15}{2}}\cdot\sqrt{\frac{10\cdot\left(a-1\right)^2}{3}}\)
\(P=\sqrt{\frac{15}{2}}.\sqrt{\frac{10.\left(a-1\right)^2}{3}}\) ( ĐK a<1 )
\(\Leftrightarrow P=\frac{\sqrt{15}}{\sqrt{2}}.\frac{\sqrt{10}.\sqrt{\left(a-1\right)^2}}{\sqrt{3}}\)
\(\Leftrightarrow P=\frac{\sqrt{15}.\sqrt{2}}{2}.\frac{\sqrt{10}.\sqrt{3}.\left|a-1\right|}{3}\)
\(\Leftrightarrow P=\frac{\sqrt{30}}{2}.\frac{\sqrt{30}\left(1-a\right)}{3}\)( vì a-1<0)
\(\Leftrightarrow P=\frac{\sqrt{30}.\sqrt{30}\left(1-a\right)}{2.3}\)
\(\Leftrightarrow\frac{30\left(1-a\right)}{6}\)
\(\Leftrightarrow5\left(1-a\right)\)
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Rút gọn
A=\(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
B=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+\sqrt{2\sqrt{5}}}}\)
\(A=\sqrt{2}\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\sqrt{2}\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right).\left(16-15\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{3}+\sqrt{5}\right).\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
Mà \(B>0\) \(\Rightarrow B=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)