CM:a)\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{4}\)
b)\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{10}\)
Tính:
A=\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
B=\(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)
C=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
D=\(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)(2 cách)
F=\(\dfrac{\sqrt{17-12\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}-\dfrac{\sqrt{17}+12\sqrt{2}}{\sqrt{3+2\sqrt{2}}}\)
\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
Cách 1:
\(E=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)
\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
Cách 2:
\(E^2=(4+\sqrt{15})^2(\sqrt{10}-\sqrt{6})^2(4-\sqrt{15})=(4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15}).(16-4\sqrt{15})\)
\(=(16-15)(4+\sqrt{15})(4-\sqrt{15}).4=(16-15)(16-15).4=4\)
Vì $E>0$ nên $E=2$
Tính
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
b) \(\sqrt{4+\sqrt{7}} -\sqrt{4-\sqrt{7}}\)
c) \(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
a: =2-căn 3-2-căn 3
=-2căn 3
b: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+1-\sqrt{7}+1\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
c: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)
\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)
=>\(A=\sqrt{10+2\sqrt{5}}\)
\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)\(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
* Thực hiện phép tính
a, A= \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
b, B= \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, C= \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)
a)\(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
b) \(\sqrt{7+4\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
c) \(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
d)\(\sqrt{7+2\sqrt{10}}-\sqrt{3-2\sqrt{2}}\)
a) \(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=2+\sqrt{3}-1-\sqrt{3}=1\)
c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)
d) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)
1) thực hiện phép tính:
a) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
giúp mk vs ah mk đang cần gấp lắm
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Rút gọn biểu thức:
a)\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
b)\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
c)\(5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}\)
d)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
e)\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
Rút gọn biểu thức:
a,\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
\(b,\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)