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\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{1}=1\)
b,c
\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)
=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
Ta có: \(A=\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\)
\(=\sqrt{2011}-\sqrt{2010}< \sqrt{2011}.\sqrt{2010}=B\)
Vậy A<B
a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)
\(\sqrt{6}< \sqrt{6,25}=2,5\);
\(\sqrt{12}< \sqrt{12,25}=3,5\);
\(\sqrt{20}< \sqrt{20,25}=4,5\)
=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)
Vậy P < 12
Answer:
ý a, tham khảo bài làm của @xyzquynhdi
\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{5}+2\sqrt{3}\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
Lời giải:
Ta có:
\(A=\sqrt{30}-\sqrt{29}=\frac{30-29}{\sqrt{30}+\sqrt{29}}=\frac{1}{\sqrt{30}+\sqrt{29}}\)
\(B=\sqrt{29}-\sqrt{28}=\frac{29-28}{\sqrt{29}+\sqrt{28}}=\frac{1}{\sqrt{29}+\sqrt{28}}\)
Mà : \(\sqrt{30}+\sqrt{29}> \sqrt{29}+\sqrt{28}\Rightarrow \frac{1}{\sqrt{30}+\sqrt{29}}< \frac{1}{\sqrt{29}+\sqrt{28}}\Rightarrow A< B\)
Lời giải :
a) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)
\(=0,1-\sqrt{0,1}\)
b) \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
c) \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
d) \(\sqrt{9-4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)
e) \(\sqrt{16-6\sqrt{7}}=\sqrt{9-2\cdot3\cdot\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7}\)
A= \(\frac{\left(\sqrt{30}\right)^2-\left(\sqrt{29}\right)^2}{\sqrt{30}+\sqrt{29}}\)= \(\frac{1}{\sqrt{30}+\sqrt{29}}\)
B= \(\frac{\left(\sqrt{29}\right)^2-\left(\sqrt{28}\right)^2}{\sqrt{29}+\sqrt{28}}\)= \(\frac{1}{\sqrt{29}+\sqrt{28}}\)
Mà ta có \(\sqrt{30}+\sqrt{29}\)>\(\sqrt{28}+\sqrt{29}\)
Nên \(\frac{1}{\sqrt{30}+\sqrt{29}}\)<\(\frac{1}{\sqrt{29}+\sqrt{28}}\)
Suy ra A<B
CÓ MA BIẾT KIT