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\(\sqrt{8-2\sqrt{15}}+\sqrt{48+6\sqrt{15}}\\ =\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{45+2\cdot3\sqrt{5}\cdot\sqrt{3}+3}\\ =\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}+\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}+3\sqrt{5}+\sqrt{3}=4\sqrt{5}\)
\(\sqrt{8-\sqrt{60}}-\sqrt{23-\sqrt{240}}\\ =\sqrt{8-\sqrt{4\cdot15}}-\sqrt{23-\sqrt{4\cdot60}}\\ =\sqrt{8-2\sqrt{15}}-\sqrt{23-2\sqrt{60}}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{20-2\cdot\sqrt{20}\cdot\sqrt{3}+3}\\ =\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}-\sqrt{20}+\sqrt{3}\\ =\sqrt{5}-2\sqrt{5}=-\sqrt{5}\)
a) Bình phương lên,ta so sánh \(\left(\sqrt{5}+\sqrt{7}\right)^2=5+2\sqrt{35}+7\text{ và }12\)
Xét hiệu hai vế \(\left(\sqrt{5}+\sqrt{7}\right)^2-12=2\sqrt{35}>0\) nên ....
b) \(14=\sqrt{14^2}=\sqrt{196}>\sqrt{195}=\sqrt{13}.\sqrt{15}\)
c) \(\left(\sqrt{8}+3\right)^2=8+2.\sqrt{72}+9;\left(6+\sqrt{2}\right)^2=36+2\sqrt{72}+2\)
\(\left(8+\sqrt{3}\right)^2-\left(6+\sqrt{2}\right)^2=\left(8+9\right)-\left(36+2\right)< 0\)
Do đó \(\left(8+\sqrt{3}\right)^2< \left(6+\sqrt{2}\right)^2\) suy ra \(\left(8+\sqrt{3}\right)< \left(6+\sqrt{2}\right)\)
d) So sánh \(\sqrt{27}+\sqrt{6}\text{ và }\sqrt{48}-1\)
Dễ chứng minh \(\sqrt{27}+\sqrt{6}> \sqrt{48}-1\)
Suy ra \(\sqrt{27}+\sqrt{6}+1>\sqrt{48}\) (thêm 1 vào mỗi vế)
a)
\(\left(\sqrt{5}+\sqrt{7}\right)^2=12+2\sqrt{35}\)
\(\sqrt{12}^2=12\)
=>\(\sqrt{5}+\sqrt{7}>\sqrt{12}\)
các câu còn lại cũng làm như vậy
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
a) = \(5\sqrt{2}-3\sqrt{6}+3\sqrt{2}+5\sqrt{6}\)
= \(8\sqrt{2}+2\sqrt{6}\)
b) = \(2\sqrt{3}-4\sqrt{2}-5\sqrt{3}-\sqrt{2}\)
= \(-3\sqrt{3}-5\sqrt{2}\)
c) = \(\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=\(\frac{2\sqrt{2}+2-2-\sqrt{2}}{2^2-\sqrt{2^2}}\)
=\(\frac{\sqrt{2}}{4-2}\) = \(\frac{\sqrt{2}}{2}\)
d) = \(2\sqrt{6}-5\sqrt{6}+2\sqrt{2}\)
=\(-3\sqrt{6}+2\sqrt{2}\)
e) = \(8\sqrt{6}+3\sqrt{6}-6\sqrt{6}=5\sqrt{6}\)
f) = \(4\sqrt{3}+9\sqrt{3}-4\sqrt{3}=9\sqrt{3}\)
g) = \(10+5\sqrt{10}-5\sqrt{10}=10\)
h) = \(\frac{\left(3+\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{\left(3-\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
= \(\frac{9+3\sqrt{3}+3\sqrt{3}+3}{3^2-\sqrt{3^2}}+\frac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-\sqrt{3^2}}\)
= \(\frac{12+6\sqrt{3}}{9-3}+\frac{12-6\sqrt{3}}{9-3}\)
= \(\frac{12+6\sqrt{3}+12-6\sqrt{3}}{6}\)
= \(\frac{24}{6}=4\)
k) = \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
l) = \(\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\)
= \(\frac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{\sqrt{8^2}-2^2}\)
= \(\frac{2\sqrt{6}}{8-4}=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\)
\(\sqrt{37}-\sqrt{15};2\)
có \(\left(\sqrt{37}-\sqrt{15}\right)^2=37-2\sqrt{555}+15=52-2\sqrt{555}\)
\(2^2=4\)
xét \(52-2\sqrt{555}-4=48-2\sqrt{555}\)
SS:\(48;2\sqrt{555}\)
\(48^2=2304\)
\(\left(2\sqrt{555}\right)^2=2220\)
2304>2220=>\(\sqrt{37}-\sqrt{15}>2\)
\(\left(\sqrt{24}-\sqrt{48}-\sqrt{6}\right)\sqrt{6}+12\sqrt{2}=\left(2\sqrt{6}-4\sqrt{3}-\sqrt{6}\right)\sqrt{6}+12\sqrt{2}\)
\(=12-12\sqrt{2}-6+12\sqrt{2}=6\)
\(\left(\sqrt{24}-\sqrt{48}-\sqrt{6}\right)\sqrt{6}+12\sqrt{2}\)
\(=\sqrt{24.6}-\sqrt{48.6}-6+12\sqrt{2}\)
\(=\sqrt{144}-\sqrt{288}-6+12\sqrt{2}\)
\(=12-12\sqrt{2}-6+12\sqrt{2}\)
\(=6\)