Chứng minh rằng: \(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{49\cdot50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
help mik với
Chứng minh
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{49\cdot50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
Lời giải:
\(\text{VT}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+....+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{49}+\frac{1}{50}\)
Do đó ta có đpcm.
Chứng minh rằng :
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\\ =1+\dfrac{1}{2}+...+\dfrac{1}{50}-1-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{25}\\ =\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
1, Chứng minh
a, \(\dfrac{1}{1\cdot2}\) + \(\dfrac{1}{3\cdot4}\) + \(\dfrac{1}{5\cdot6}\) + .... + \(\dfrac{1}{49\cdot50}\) = \(\dfrac{1}{26}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{28}\) +.... + \(\dfrac{1}{50}\)
\(LINH_1=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+....+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=LINH_2\left(đpcm\right)\)
Cho M= \(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+....+\dfrac{1}{99\cdot100}\)
Chứng minh rằng: \(\dfrac{7}{12}< M< \dfrac{5}{6}\)
Chứng minh rằng : \(\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\)
Ta có :
Vế phải =1 - 1/2 + 1/3 - 1/4 + ... + 1/49 - 1/50
= (1+ 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + ... +1/50)
<=> (1 + 1/2 + 1/3 + 1/4 + ... + 1/49+1/50)- 2(1/2 +1/4 +...+1/50)
=(1+1/2 +1/3 +1/4...+ 1/49+1/50) - (1+1/2 +...+1/25)
=1/26 + 1/27 +1/28 +...+1/50 (đpcm)
A=\(\dfrac{1}{1\cdot2}\)+\(\dfrac{1}{2\cdot3}\)+\(\dfrac{1}{3\cdot4}\)+...+\(\dfrac{1}{49\cdot50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
Ta có :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
~ Chúc bn học tốt ~
11⋅2" id="MathJax-Element-1-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">12⋅3" id="MathJax-Element-2-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">+13⋅4" id="MathJax-Element-3-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">+...+149⋅50" id="MathJax-Element-4-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" tabindex="0">
+\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=1-\dfrac{1}{50}\)
\(A=\dfrac{49}{50}\)
\(B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+\dfrac{2}{6\cdot7\cdot8}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)
Chứng tỏ rằng \(\dfrac{A}{B}\) là một số nguyên biết :
\(B=\dfrac{1}{1008\cdot2014}+\dfrac{1}{1009\cdot2013}+\dfrac{1}{1010\cdot2012}+...+\dfrac{1}{2014\cdot1008}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{2013\cdot2014}\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
1/* Chứng minh rằng:
\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+..+\dfrac{1}{50}\)
2/* Cho:
A=\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+.....+\dfrac{1}{99\times100}\). Chứng minh rằng:\(\dfrac{7}{12}< A>\dfrac{5}{6}\)
Các bn giúp mk những bài này nha!
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
1/ Ta có :
\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{49\times50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{50}\right)\)
\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\times2\)
\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)
\(\Rightarrow\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}\)
Hay \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...+\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
~ Học tốt nha ~