Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
Những câu hỏi liên quan
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
chứng tỏ rằng :
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3.\left(2-1\right)}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5\left(3-2\right)}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011\left(2006-2005\right)}\)
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011}\)
Nhận xét: (a-b)2 \(\ge\) 0 => a2 + b2 \(\ge\) 2ab
Áp dụng ta có: \(3=\left(\sqrt{2}\right)^2+\left(\sqrt{1}\right)^2\ge2.\sqrt{2}.\sqrt{1}\)
\(5=\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2\ge2.\sqrt{3}.\sqrt{2}\)
...
\(4011=\left(\sqrt{2006}\right)^2+\left(\sqrt{2005}\right)^2\ge2.\sqrt{2006}.\sqrt{2005}\)
=> \(VT
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Chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}< \frac{3}{7}\)
Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)
a/Chứng minh rằng \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b/Áp dụng chứng minh
\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)
Rút gọn biểu thức 1) frac{sqrt{5+2sqrt{6}}+sqrt{8+2sqrt{15}}}{sqrt{7+2sqrt{10}}}2) left(2+frac{3+sqrt{3}}{sqrt{3}+1}right)left(2+frac{3-sqrt{3}}{sqrt{3}-1}right):left(sqrt{5}-2right)3) left(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)4) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+...+frac{1}{sqrt{99}+sqrt{100}}5) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...-frac{1}{sqrt{98}-sqrt{99}}+frac{1}{sqrt{99}-s...
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Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
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Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
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Sao làm hổng ai bảo đú.n/g vậy :(((
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Mình rút gọn như thế này đúng không nhỉ?Pleft(2-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{2x-sqrt{x}-3}+frac{sqrt{x}}{sqrt{x}+1}right)Pleft[frac{2left(2sqrt{x}-3right)}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right]:left[frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}+frac{sqrt{x}left(2sqrt{x}-3right)}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}right]Pleft(frac{4sqrt{x}-6}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x...
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Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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Tính:
a) Asqrt{8-2sqrt{15}}left(sqrt{3}+sqrt{5}right)-left(sqrt{45}-sqrt{20}right)
b) Bleft(frac{sqrt{21}-sqrt{3}}{sqrt{7}-1}-frac{sqrt{15}-sqrt{3}}{1-sqrt{5}}right)left(frac{1}{2}sqrt{6}-sqrt{frac{3}{2}}+3sqrt{frac{2}{3}}right)
c) C2sqrt{3}+sqrt{7-4sqrt{3}}+left(sqrt{frac{1}{3}}-sqrt{frac{4}{3}+}sqrt{3}right):sqrt{3}
d) Dleft(frac{5+sqrt{5}}{5-sqrt{5}}+frac{5-sqrt{5}}{5+sqrt{5}}right):frac{1}{sqrt{7-4sqrt{3}}}
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Tính:
a) \(A=\sqrt{8-2\sqrt{15}}\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
b) \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
c) \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}+}\sqrt{3}\right):\sqrt{3}\)
d) \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
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b, MA-Bfrac{sqrt{x}+2}{sqrt{x}+3}-left(frac{5}{x+sqrt{x}-6}+frac{1}{sqrt{x}-2}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}-frac{1}{sqrt{x}-2}frac{left(sqrt{x}+2right)left(sqrt{x}-2right)}{x+sqrt{x}-6}-frac{5}{x+sqrt{x}-6}-frac{1left(sqrt{x}+3right)}{x+sqrt{x}-6}frac{x-4-5-sqrt{x}-3}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-4sqrt{x}+3sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{left(sqrt{x}-4right)left(sqrt{x...
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b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
bạn trung học hay tiểu học vậy