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a. \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
= \(3-\sqrt{6} +2\sqrt{6}-3\) = \(\sqrt{6}\)
b. \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
= \(\sqrt{8\sqrt{3}}-2.5\sqrt{12}+4\sqrt{8\sqrt{3}}\)
= \(5\sqrt{8\sqrt{3}}-5\sqrt{4.\sqrt{12}}=5\sqrt{8\sqrt{3}}-5\sqrt{4.2\sqrt{3}}\)
= \(5\sqrt{8\sqrt{3}}-5\sqrt{8\sqrt{3}}=0\)
c. \(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\) = \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}.\left(\sqrt{3}+1\right)\)
=\(\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
= 3 - 1 = 2
d. \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)}{\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\) = \(\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}\)
= \(\dfrac{2\sqrt{5}}{\sqrt{2}}\)= \(\sqrt{10}\)
e. \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right)\)\(2.\left(3+2\sqrt{2}+2-1+3-2\sqrt{2}\right)=2.7=14\)
a: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)
a: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2\left(\sqrt{5}+1\right)\)
\(=2\sqrt{5}-2\sqrt{5}-2=-2\)
c: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{6}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{6}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)=2\sqrt{2}\)
b) Tương tự
b) \(\sqrt{7-2\sqrt{10}}\) - \(\sqrt{7+2\sqrt{10}}\)
= \(\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{2}+\left(\sqrt{2}\right)^2}\) - \(\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
= \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\) - \(\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
= \(\left(\sqrt{5}-\sqrt{2}\right)\) - \(\left(\sqrt{5}+\sqrt{2}\right)\)
= \(\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
= \(-2\sqrt{2}\)
1a) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{8-4\sqrt{2}-\sqrt{16}+2\sqrt{8}}\)
\(=\sqrt{8-4\sqrt{2}-4+4\sqrt{2}}\)
\(=\sqrt{4}=2\)
1b) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+4\sqrt{3}+3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
a/ \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\) \(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\).
b/ \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\Rightarrow A^2=8+2\sqrt{4^2-\left(\sqrt{10+2\sqrt{5}}\right)^2}=8+2\sqrt{6-2\sqrt{5}}\) \(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{5}+1\)
c/ \(B=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\Rightarrow\sqrt{2}B=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2=2\Rightarrow B=\sqrt{2}\)
a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|=\left(\sqrt{5}-\sqrt{2}\right)-\left(\sqrt{5}+\sqrt{2}\right)\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}=-2\sqrt{2}\)
c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|=\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
d) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=2\sqrt{6+2\sqrt{5}}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{5}-2=2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
e,
\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{4}\cdot\sqrt{2}}+\sqrt{9+2\cdot\sqrt{4}\cdot\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{8}}+\sqrt{9+2\cdot\sqrt{8}}\\ =\sqrt{9-2\cdot3\cdot\sqrt{8}+8}+\sqrt{8+2\cdot\sqrt{8}\cdot1+1}\\ =\sqrt{3^2-2\cdot3\cdot\sqrt{8}+\sqrt{8}^2}+\sqrt{\sqrt{8}^2+2\cdot\sqrt{8}\cdot1+1^2}\\ =\sqrt{\left(3-\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{8}+1\right)^2}\\ =\left|3-\sqrt{8}\right|+\left|\sqrt{8}+1\right|\\ =3-\sqrt{8}+\sqrt{8}+1\\ =4\)
f,
\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{9}\cdot\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{18}}\\ =\sqrt{4-2\cdot2\cdot\sqrt{2}+2}+\sqrt{18-2\cdot2\cdot\sqrt{18}+4}\\ =\sqrt{2^2-2\cdot2\cdot\sqrt{2}+\sqrt{2}^2}+\sqrt{\sqrt{18}^2-2\cdot2\cdot\sqrt{18}+2^2}\\ =\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{18}-2\right)^2}\\ =\left|2-\sqrt{2}\right|+\left|\sqrt{18}-2\right|\\ =2-\sqrt{2}+\sqrt{18}-2\\ =\sqrt{2}+\sqrt{18}\\ =\sqrt{2}+\sqrt{9}\cdot\sqrt{2}\\ =\sqrt{2}+3\sqrt{2}\\ =4\sqrt{2}\)
e ; f \(\Rightarrow\) bó tay
sao ko có e,f vậy bn
rứa thôi, cảm ơn bạn
cảm ơn bạn nhiều nha~~
a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
Đặt \(A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(\Rightarrow\sqrt{2}A=\sqrt{2}\left(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\right)\)
\(\sqrt{2}A=\sqrt{10+4\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(\sqrt{2}A=\sqrt{\left(2+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(\sqrt{2}A=\left|2+\sqrt{6}\right|-\left|\sqrt{6}-2\right|\)
\(\sqrt{2}A=\sqrt{6}+2-\sqrt{6}+2\)
\(\sqrt{2}A=4\)
\(\Rightarrow A=2\sqrt{2}\)
b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
Đặt \(B=\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(\Rightarrow\sqrt{2}B=\sqrt{2}\left(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\right)\)
\(=\sqrt{14-4\sqrt{10}}-\sqrt{14+4\sqrt{10}}\)
\(=\sqrt{\left(\sqrt{10}-2\right)^2}-\sqrt{\left(\sqrt{10}+2\right)^2}\)
\(=\left|\sqrt{10}-2\right|-\left|\sqrt{10}+2\right|\)
\(=\sqrt{10}-2-\sqrt{10}-2\)
\(=-4\)
\(\Rightarrow B=\dfrac{-4}{\sqrt{2}}=-2\sqrt{2}\)
c)\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3}+1\)
\(=2\sqrt{3}\)
f) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)
\(=\left|2-\sqrt{2}\right|+\left|3\sqrt{2}-2\right|\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=2\sqrt{2}\)
uk , e, f khó thật
f ) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)
=\(\sqrt{4-4\sqrt{2}+2}+\sqrt{22-4.\sqrt{9}\sqrt{2}}\)
= \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{18-4\sqrt{18}+4}\)
= \(\left|2-\sqrt{2}\right|+\sqrt{\left(\sqrt{18}-2\right)^2}\)
= \(2-\sqrt{2}+\left|3\sqrt{2}-2\right|\)
= \(2-\sqrt{2}+3\sqrt{2}-2\)
= \(2\sqrt{2}\)
có gì đó sai sai !!!!!!!!!!!
\(-\sqrt{2}=\sqrt{2}\)
Mình sẽ giải câu e và f
e) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.1+1}=\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.3+3^2}+\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.1+1}=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=\left|2\sqrt{2}-3\right|+\left|2\sqrt{2}+1\right|=3-2\sqrt{2}+2\sqrt{2}+1=4\)
f) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}=\sqrt{4-2.2\sqrt{2}+2}+\sqrt{18-2.3\sqrt{2}.2+4}=\sqrt{2^2-2.2\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}\right)^2-2.3\sqrt{2}.2+2^2}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=\left|2-\sqrt{2}\right|+\left|3\sqrt{2}-2\right|=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)
Kết quả là \(2\sqrt{2}\) mà bạn