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a) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left(\sqrt{5}-2\right)-\left(\sqrt{5}+2\right)=-4\)
b) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\frac{1}{\sqrt{2}}.\left(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{7}-1-\sqrt{7}-1\right)=-\sqrt{2}\)
c) \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)
\(=7-3\sqrt{5}-\left(7+3\sqrt{5}\right)=-6\sqrt{5}\)
\(a.\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{49-2.7.3\sqrt{5}+45}-\sqrt{49+2.7.3\sqrt{5}+45}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\) \(b.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\) \(c.\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
\(a,\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{8+2.2\sqrt{2}.5+25}-\sqrt{2-2.\sqrt{2}.3+9}\)
\(=\sqrt{\left[2\sqrt{2}+5\right]^2}-\sqrt{\left[\sqrt{2}-3\right]^2}\)
\(=2\sqrt{2}+5-\left(3-\sqrt{2}\right)\)
\(=2+\sqrt{2}\)
chúc bn học tốt
a) \(\sqrt{\left(2\sqrt{2}+5\right)^2}\) \(-\) \(\sqrt{\left(3-\sqrt{2}\right)^2}\)= \(|2\sqrt{2}+5|\)\(-\)\(|3-\sqrt{2}|\)
\(=\)\(2\sqrt{2}+5-3+\sqrt{2}=2+3\sqrt{2}\)
b)\(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)
a/ \(=\sqrt{\left(5+2\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=5+2\sqrt{2}-3+\sqrt{2}=2+3\sqrt{2}\)
b/ \(=\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\)
a)
\(\sqrt{33+20\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{33+2\sqrt{200}}-\sqrt{11-2\sqrt{18}}\\ =\sqrt{\sqrt{8^2}+2\sqrt{8}\sqrt{25}+5^2}-\sqrt{\sqrt{2^2}-2\sqrt{2}\sqrt{9}+3^2}\\ =\sqrt{\left(\sqrt{8}+5\right)^2}-\sqrt{\left(\sqrt{2}-3\right)^2}\\ =\left|\sqrt{8}+5\right|-\left|\sqrt{2}-3\right|\\ =\sqrt{8}+5-3+\sqrt{2}\\ =3\sqrt{2}+2\)
b)
\(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\\ =\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\\ =\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\\ =7-\sqrt{45}-7-\sqrt{45}\\ =-2\sqrt{45}\)
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) Ta có: \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\left|\sqrt{10}-1\right|=\sqrt{10}-1\)
b) Ta có: \(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|\)
\(=\sqrt{7}-\sqrt{2}\)
c) Ta có: \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1+\sqrt{3}-1\)
\(=2\sqrt{3}\)
d) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
e) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
g) Ta có: \(\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|3+\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}+3+\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(=3+2\sqrt{3}+2\sqrt{2}\)
h) Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{\left(\sqrt{3}+2\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\cdot\left(\sqrt{3}+2\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
k) Ta có: \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}\)
\(=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\)
\(=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\)
\(=7-\sqrt{45}-\left(7+\sqrt{45}\right)\)
\(=7-\sqrt{45}-7-\sqrt{45}\)
\(=-2\sqrt{45}=-6\sqrt{5}\)
i) Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\cdot\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)
\(=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\cdot\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow A=\sqrt{5}+1\)
\(\left(\sqrt{a+42}+\sqrt{a-42}\right)^2=16\)
<=> a+\(\sqrt{a^2-42^2}=8\)
<=>\(a-8=\sqrt{a^2-42^2}\)
<=>\(\left(a-8\right)^2=a^2-42^2\)
<=>16a-1828=0
<=>a=\(\frac{457}{4}\)
thay vaof bt kia nha banj