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13 tháng 7 2017
a. \(VT=\sqrt{14+2\sqrt{13}}-\sqrt{14-2\sqrt{13}}\)
=\(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{\left(\sqrt{13}-1\right)^2}=\sqrt{13}+1-\left(\sqrt{13}-1\right)\)
\(=\sqrt{13}+1-\sqrt{13}+1=2=VP\left(đpcm\right)\)
b. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{5-2\sqrt{6}}-\sqrt{2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{2}\)
\(=2+\sqrt{3}-\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{2}=2+\sqrt{3}-\sqrt{3}+\sqrt{2}-\sqrt{2}\)
\(=2=VP\left(đpcm\right)\)
TH
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}-\sqrt{12+2\sqrt{11}}}\right)\left(\sqrt{11}+\sqrt{3}\right)\)
3
5 tháng 10 2020
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{\left(\sqrt{13+1}\right)^2}}-\sqrt{\left(\sqrt{11+1}\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{13+2}}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)\(=\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{11}+\sqrt{13}\right)=13-11=2\)
M
0
26 tháng 7 2020
Trả lời:
\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)
\(A=\frac{\sqrt{2}.\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)
\(A=\frac{\sqrt{2}.\sqrt{4+2\sqrt{3}}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)
\(A=\frac{\sqrt{2}.\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)
\(A=\frac{\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)
\(A=\frac{\sqrt{2}.\left(\sqrt{3}+1\right)}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)
\(A=1\)
22 tháng 6 2017
C = \(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{12+2\sqrt{\left(\sqrt{13}+1\right)^2}}-\sqrt{\left(\sqrt{11}+1\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{14+2\sqrt{13}}-\left(\sqrt{11}+1\right)\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
C = \(\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{13}+\sqrt{11}\right)\)
C \(\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{13}+\sqrt{11}\right)\) = \(13-11\) = \(2\)
\(\sqrt{14}-\sqrt{13}< 2\sqrt{3}-\sqrt{11}\)
\(\Leftrightarrow\sqrt{14}-\sqrt{13}< \sqrt{12}-\sqrt{11}\)
\(\Leftrightarrow\sqrt{14}+\sqrt{11}< \sqrt{12}+\sqrt{13}\)
\(\Leftrightarrow14+11+2\sqrt{14.11}< 12+13+2\sqrt{12.13}\)
\(\Leftrightarrow25+2\sqrt{154}< 25+2\sqrt{156}\)
\(\Leftrightarrow\sqrt{154}< \sqrt{156}\)(luôn đúng)
Vậy \(\sqrt{14}-\sqrt{13}< 2\sqrt{3}-\sqrt{11}\)