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a) \(\sqrt{8-\sqrt{60}}\)=\(\sqrt{8-\sqrt{4.15}}\)=\(\sqrt{8-2\sqrt{15}}\)=\(\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}\)=\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)=l\(\sqrt{5}\)\(-\sqrt{3}\)l =\(\sqrt{5}\)\(-\sqrt{3}\)(do \(\sqrt{5}\)\(-\sqrt{3}\)>0)
a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{3-2\sqrt{3}+1}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{9+6\sqrt{2}+2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
c) \(\sqrt{25x^2}-2x=-5x-2x=-7x\)(vì x < 0)
d) \(x-5+\sqrt{25-10x+x^2}=x-5+\sqrt{\left(5-x\right)^2}=x-5+x-5=2x-10\) (vì x > 5)
a) Ta có: \(\sqrt{16-6\sqrt{7}}+\sqrt{7}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+7}+\sqrt{7}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{7}\)
\(=\left|3-\sqrt{7}\right|+\sqrt{7}\)
\(=3-\sqrt{7}+\sqrt{7}\)
\(=3\)
b) Ta có: \(\sqrt{\left|12\sqrt{5}-29\right|}+\sqrt{12\sqrt{5}+29}\)
\(=\sqrt{\left(\sqrt{29-12\sqrt{5}}+\sqrt{12\sqrt{5}+29}\right)^2}\)
\(=\sqrt{29-12\sqrt{5}+2\sqrt{\left(29-12\sqrt{5}\right)\left(12\sqrt{5}+29\right)}+12\sqrt{5}+29}\)
\(=\sqrt{58+2\sqrt{121}}\)
\(=\sqrt{58+2.11}\)
\(=\sqrt{80}=4\sqrt{5}\)
\(\text{Đặt }A=\sqrt{4+\sqrt{10-2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10-2\sqrt{5}}\)
\(+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{16-10+2\sqrt{5}}=8+2\sqrt{6+2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}+1\right)^2}=8+2\left(\sqrt{5}+1\right)\)
\(=10+2\sqrt{5}\)
\(\Rightarrow A=\sqrt{10-2\sqrt{5}}\text{Hoặc }A=-\sqrt{10-2\sqrt{5}}\)
\(\text{Mà }A>0\text{ nên: }A=\sqrt{10-2\sqrt{5}}\)
a) \(A=\frac{1}{2}\sqrt{32}+\sqrt{98}-\frac{1}{6}\sqrt{18}=\frac{1}{2}\sqrt{4^2.2}+\sqrt{7^2.2}-\frac{1}{6}.\sqrt{3^2.2}\)
\(=\frac{1}{2}\sqrt{4^2}.\sqrt{2}+\sqrt{7^2}.\sqrt{2}-\frac{1}{6}.\sqrt{3^2}.\sqrt{2}\)\(=\frac{1}{2}.4\sqrt{2}+7\sqrt{2}-\frac{1}{6}.3.\sqrt{2}\)\(=2.\sqrt{2}+7\sqrt{2}-\frac{1}{2}\sqrt{2}=\left(2+7-\frac{1}{2}\right)\sqrt{2}=\frac{17}{2}\sqrt{2}\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)
=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)
12\(\sqrt{3}\)
à
a)\(\sqrt{\left(\sqrt{20}\right)^2-2.\sqrt{20}.\sqrt{9}+\left(\sqrt{9}\right)^2}=\sqrt{\left(\sqrt{20}-\sqrt{9}\right)^2}=\left|\sqrt{20}-\sqrt{9}\right|=\sqrt{20}-3=2\sqrt{5}-3\)
b)\(\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}\)
c)\(\sqrt{5-2\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left|\sqrt{5}-\sqrt{2}\right|=\sqrt{5}-\sqrt{2}\)
d)\(\sqrt{12+2.\sqrt{12}.\sqrt{5}+5}=\sqrt{\left(\sqrt{12}+\sqrt{5}\right)^2}=\left|\sqrt{12}+\sqrt{5}\right|=\sqrt{12}+\sqrt{5}=2\sqrt{3}+\sqrt{5}\)
e)\(\sqrt{18-2.3\sqrt{2}.1+1}=\sqrt{\left(3\sqrt{2}-1\right)^2}=\left|3\sqrt{2}-1\right|=3\sqrt{2}-1\)
h) \(\sqrt{12+2.\sqrt{12}.\sqrt{9}+9}=\sqrt{\left(\sqrt{12}+\sqrt{9}\right)^2}=\left|\sqrt{12}+\sqrt{9}\right|=\sqrt{12}+\sqrt{9}=2\sqrt{3}+3\)