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đk n khác 0 để psố có nghĩa
qui đồng ta có;
4 +2m = 2n
2(n-m) =4
n-m =2
nếu n = 1 thi m =-1
nếu n =2 thì m=0
.....
đk 0 khác 0 để psố coc nghĩa
quy đồng ta có
4 + 2m = 2n
2(n-m)=4
n-m=2
nếu n=1 thì m= 1
nếu n=2 thì m=0
ai thấy đúng thì tích mk nha
A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được:
\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)
=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)
\(4A>1=>A>\frac{1}{4}\)
Sao k có ai giúp mk hết vậy >:((, thôi để mk tự giúp mk vậy :>. E mới nghĩ ra cách này có gì sai anh giúp đỡ.
Cách 1 - Ta có :
\(A=\frac{1}{1.2}+\frac{1}{1.3}+\frac{1}{1.4}+...+\frac{1}{3.2}+\frac{1}{3.3}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{5}{6}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
Mà \(\frac{5}{6}>\frac{2}{3}\Rightarrow\frac{5}{6}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}>\frac{2}{3}\)
\(\Leftrightarrowđpcm\)
\(\frac{1}{1.3}+\left|\frac{-1}{3.5}\right|+\left|\frac{-1}{5.7}\right|+...+\left|\frac{-1}{49.51}\right|=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\left(1-\frac{1}{51}\right)=\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)
\(\frac{1}{1.3}+\left|-\frac{1}{3.5}\right|+\left|-\frac{1}{5.7}\right|+...+\left|-\frac{1}{49.51}\right|\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49\cdot51}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\frac{50}{51}\)
\(=\frac{25}{51}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\frac{1}{1.3}+\frac{1}{3,5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{101-99}{99.101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}\)
\(=\frac{50}{101}\)
-.-
\(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\)
\(=\frac{2}{2}.\left(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\right)\)
\(=\frac{2}{1.1.2}+\frac{2}{1.3.2}+\frac{2}{3.2.2}+...+\frac{2}{49.25.2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=2.\left(1-\frac{1}{50}\right)\)
\(=2.\frac{49}{50}\)
\(=\frac{49}{25}\)
Chúc bạn học tốt !!!