GH
2
K
Khách
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13 tháng 8 2020
Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=21\cdot\left[2+\sqrt{3}+3-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right]-6\cdot\left[2-\sqrt{3}+3+\sqrt{5}+2\cdot\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{\left(4+2\sqrt{3}\right)\left(6-2\sqrt{5}\right)}\right)-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\sqrt{\left(4-2\sqrt{3}\right)\left(6+2\sqrt{5}\right)}\right]-15\sqrt{15}\)
\(=21\cdot\left[5+\sqrt{3}-\sqrt{5}+\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\right]-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\left(\sqrt{3}-1\right)\left(\sqrt{5}+1\right)\right]-15\sqrt{15}\)
\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{15}-\sqrt{3}+\sqrt{5}-1\right)-6\cdot\left(5-\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt{3}-\sqrt{5}-1\right)-15\sqrt{15}\)
\(=21\cdot\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)
\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)
\(=60\)
14 tháng 8 2019
\(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=1-\sqrt{3}-\sqrt{3}-2\)
\(=-2\sqrt{3}-1\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+4-2\sqrt{3}\)
\(=6-3\sqrt{3}\)
ST
3
KT
30 tháng 7 2018
\(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(\Rightarrow\)\(\sqrt{2}A=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1+\sqrt{3}-1\)
\(=2\sqrt{3}\)
\(\Rightarrow\)\(A=\sqrt{6}\) (đpcm)
30 tháng 7 2018
\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
\(VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\sqrt{\frac{2\left(2+\sqrt{3}\right)}{2}}+\sqrt{\frac{2\left(2-\sqrt{3}\right)}{2}}\)
\(=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\)
\(=\sqrt{\frac{3+2\sqrt{3}+1}{2}}+\sqrt{\frac{3-2\sqrt{3}+1}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{3}+\sqrt{1}\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-\sqrt{1}\right)^2}{2}}\)
\(=\frac{\left|\sqrt{3}+\sqrt{1}\right|+|\sqrt{3}-\sqrt{1}|}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{1}+\sqrt{3}-\sqrt{1}}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{12}}{\sqrt{2}}=\sqrt{6}\)
\(=VP\)
Vậy đẳng thức được chứng minh .
BH
15 tháng 8 2018
\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)
\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)
KH
0
26 tháng 8
a: \(\sqrt{6-4\sqrt2}+\sqrt{22-12\sqrt2}\)
\(=\sqrt{4-2\cdot2\cdot\sqrt2+2}+\sqrt{18-2\cdot3\sqrt2\cdot2+4}\)
\(=\sqrt{\left(2-\sqrt2\right)^2}+\sqrt{\left(3\sqrt2-2\right)^2}\)
\(=2-\sqrt2+3\sqrt2-2=2\sqrt2\)
b: \(\sqrt{\left(\sqrt3-\sqrt2\right)^2}+\sqrt2=\sqrt3-\sqrt2+\sqrt2=\sqrt3\)
c: \(3\sqrt5-\sqrt{\left(1-\sqrt5\right)^2}\)
\(=3\sqrt5-\left|1-\sqrt5\right|\)
\(=3\sqrt5-\left(\sqrt5-1\right)=2\sqrt5+1\)
d:Sửa đề: \(\sqrt{17-12\sqrt2}+\sqrt{6+4\sqrt2}\)
\(=\sqrt{9-2\cdot3\cdot2\sqrt2+8}+\sqrt{4+2\cdot2\cdot\sqrt2+2}\)
\(=\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{\left(2+\sqrt2\right)^2}=3-2\sqrt2+2+\sqrt2=5-\sqrt2\)
2 tháng 9 2018
\(\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6+\sqrt{2}}\right)}=2\)
=2.
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 6 2019
Chắc lớp 9 thì chỉ học góc nhọn do các hệ thức lượng đều áp dụng trong tam giác vuông
\(cot\left(2a+30^0\right)=tan\left(90-2a-30^0\right)=tan\left(60^0-2a\right)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}=tan30^0\)
\(\Rightarrow60^0-2a=30^0\Rightarrow a=15^0\)
1
\(\dfrac{3-2\sqrt{3}}{\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}=\dfrac{\sqrt{3}.\left(\sqrt{3}-2\right)}{\sqrt{3}}+\dfrac{6.\left(3-\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(=\sqrt{3}-2+\dfrac{6.\left(3-\sqrt{3}\right)}{6}\)
\(=\sqrt{3}-2+3-\sqrt{3}\)
\(=1\)