tính : \(\sqrt{15-10\sqrt{11}}\) - \(\sqrt{20-6\sqrt{11}}\)
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H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
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) \(\sqrt{8+\sqrt{15}}\cdot\sqrt{8-\sqrt{15}}\)
\(=\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}\)
\(=\sqrt{8^2-\left(\sqrt{15}\right)^2}\)
\(=\sqrt{64-15}\)
\(=\sqrt{49}\)
\(=\sqrt{7^2}\)
\(=7\)
24 tháng 8 2023
a: \(=\sqrt{6^2-11}=\sqrt{25}=5\)
b: \(=\sqrt{8^2-15}=\sqrt{49}=7\)
27 tháng 7 2023
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
25 tháng 9 2021
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
27 tháng 10 2019
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
13 tháng 7 2016
@.@ Trời ơi, nhiều thế ^^
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)\)
\(=\left(\sqrt{2}.\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\frac{3\sqrt{2}}{\sqrt{5}}\right)=2\sqrt{5}-2-6+\frac{6}{\sqrt{5}}=\frac{16\sqrt{5}}{5}-8\)
b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}=\frac{75\sqrt{2}+50\sqrt{2}-45\sqrt{2}}{\sqrt{10}}=\frac{80\sqrt{2}}{\sqrt{10}}=\frac{80}{\sqrt{5}}=16\sqrt{5}\)c) \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)
\(=2+\sqrt{2}+2-\sqrt{2}=4\)
d) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)}^2\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
e) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
f)\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=1+\sqrt{2}-\sqrt{2}+1=2\)g) \(\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
3 tháng 10 2021
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)
MV
13 tháng 7 2017
a,
\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\\ =\sqrt{3-2\cdot1\cdot\sqrt{3}+1}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot1\cdot\sqrt{3}+1^2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}\\ =-1\)
b,
\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-3+\sqrt{2}\\ =\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)
c,
\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}\\ =\sqrt{5+2\cdot\sqrt{2\cdot5}+2}-\sqrt{5-2\cdot\sqrt{2\cdot5}+2}\\ =\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}\\ =2\sqrt{2}\)
d,
\(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\\ =\left(20\cdot\sqrt{20}\cdot\sqrt{15}-15\cdot\sqrt{45}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\left(20\cdot2\cdot\sqrt{5}\cdot\sqrt{15}-15\cdot3\cdot\sqrt{5}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\sqrt{15}\cdot\left(20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\right):\sqrt{15}\\ =20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\\ =40\sqrt{5}-45\sqrt{5}+5\sqrt{5}\\ =0\)
Không thể tính được do \(15< 10\sqrt{11}\)