\(\sqrt{2\sqrt{3\sqrt{4\sqrt{...\sqrt{2000}}}}}< 3\)
c/m:\(\sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{6\sqrt{7\sqrt{8.....\sqrt{1999\sqrt{2000}}}}}}}}}< 3\)
Có cách giải nhưng t ko chắc đâu nhá;) đã bảo đưa dạng a, b, c rồi mà cứ đưa dạng này-_-
\(VT< \sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{6....}}}}}=x>0\) (vô hạn dấu căn). Ta sẽ chứng minh x < 3
Ta thấy \(x^2=\sqrt{2}.x\Rightarrow x\left(x-\sqrt{2}\right)=0\Rightarrow x=\sqrt{2}< 3\Rightarrow\text{đpcm }\)
\(x^2=2\sqrt{3\sqrt{4\sqrt{5....\sqrt{2000}}}}ma?\)
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hok tốt
\(\sqrt{2\sqrt{3\sqrt{4\sqrt{5...\sqrt{1998\sqrt{1999\sqrt{2000}}}}}}}\) nhỏ hơn 3
CMR: \(\sqrt{2\sqrt{3\sqrt{4\sqrt{.....\sqrt{2000}}}}}< 3\)
\(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2000}}}}=\sqrt{2\sqrt{3\sqrt{4...\sqrt{1999\sqrt{2000}}}}}\)
\(< \sqrt{2\sqrt{3\sqrt{4...\sqrt{1999.2001}}}}< \sqrt{2\sqrt{3\sqrt{4...\sqrt{1998.\frac{1999+2001}{2}}}}}\)
\(< \sqrt{2\sqrt{3\sqrt{4...\sqrt{1998.2000}}}}< ...< \sqrt{2.\frac{3+5}{2}}\)
\(=\sqrt{2.4}=\sqrt{8}< 3\)
So sánh\(\sqrt{2\sqrt{3\sqrt{4\sqrt{...\sqrt{2000}}}}}\) và 3
CMR: \(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2000}}}}< 3\)
\(\sqrt{2\sqrt{3\sqrt{4...\sqrt{1999\sqrt{2000}}}}}< \sqrt{2\sqrt{3\sqrt{4...\sqrt{1999.2001}}}}\)
\(< \sqrt{2\sqrt{3\sqrt{4...\sqrt{1998.2000}}}}< ...< \sqrt{2.4}< 3\)
Chứng minh rằng :\(\sqrt{2\sqrt{3\sqrt{4\sqrt{5...\sqrt{2000}}}}}\)<3
Chứng minh rằng
\(P=\sqrt{2\sqrt{3\sqrt{4\sqrt{...\sqrt{2000}}}}}< 3\)
CMR:\(\sqrt{2\sqrt{3\sqrt{4\sqrt{5...\sqrt{2000}}}}}\)<3
cmr: \(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2000}}}}< 3.\)
Ta có:
\(\sqrt{2\sqrt{3\sqrt{4...\sqrt{2000}}}}\)
\(< \sqrt{2\sqrt{3\sqrt{4...\sqrt{2000.2002}}}}\)
\(=\sqrt{2\sqrt{3\sqrt{4...\sqrt{1999\sqrt{2001^2-1}}}}}\)
\(< \sqrt{2\sqrt{3\sqrt{4...\sqrt{1999.2001}}}}\)
\(........................................\)
\(< \sqrt{2.4}=\sqrt{8}< 3\)
Ta có:
√2√3√4...√2000
<√2√3√4...√2000.2002
=√2√3√4...√1999√20012−1
<√2√3√4...√1999.2001
........................................
<√2.4=√8<3