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\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\frac{36}{5\cdot7\cdot9}+...+\frac{36}{25\cdot27\cdot29}\)
\(=9\left[\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{25\cdot27\cdot29}\right]\)
\(=9\left[\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right]\)
\(=9\left[\frac{1}{3}-\frac{1}{783}\right]=9\cdot\frac{260}{783}=\frac{260}{87}\)
Đặt \(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{1}{1.3}-\frac{1}{27.29}\)
\(\Rightarrow\frac{1}{9}A=\frac{261}{783}-\frac{1}{783}\)
\(\Rightarrow\frac{1}{9}A=\frac{260}{783}\)
\(\Rightarrow A=\frac{260}{783}\div\frac{1}{9}\)
\(\Rightarrow A=\frac{2340}{783}=\frac{260}{87}\)
Bài làm:
Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)
=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)
Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)
Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)
1/21+1/28+1/36+...+2/[x.(x+1)]=2/9
=>2/42+2/56+2/72+...+2/[x.(x+1)]=2/9
=>2.(1/42+1/56+1/72+...+1/[x.(x+1)])=2/9
=>2.(1/6-1/7+1/7-1/8+1/8-1/9+...+1/x-1/(x+1))=2/9
=>1/6-1/(x+1)=1/9
=>1/(x+1)=1/18
=>x+1=18
=>x=17
Đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{21.2}+\frac{1}{28.2}+\frac{1}{36.2}...+\frac{2}{x.\left(x+1\right).2}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6.4}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}-\frac{1}{x+1}\)
\(\Rightarrow A=\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}\)
Theo bài ra ta có :
\(\left(\frac{1}{6}-\frac{1}{x+1}\right):\frac{1}{2}=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}.\frac{1}{2}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{3}{18}-\frac{2}{18}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
Vậy x = 17
a,1/5+2/5+3/5+4/5+...+9/5
=(1+2+3+4+...+9)/5
=45/5
=9
b,17,8(3,7+5,7)-7,8(4,6+4,8)
=17,8.9,4-7,8.9,4
=9,4(17,8-7,8)
=9,4.10
=94
\(a,\frac{15}{2}-\left(\frac{x}{2}-\frac{3}{4}\right)=\frac{5}{26}\)
\(\frac{x}{2}-\frac{3}{4}=\frac{15}{2}-\frac{5}{26}\)
\(\frac{x}{2}-\frac{3}{4}=39\)
\(\frac{x}{2}=39+\frac{3}{4}\)
\(\frac{x}{2}=\frac{159}{4}\)
\(\Rightarrow\frac{2.x}{4}=\frac{159}{4}\)
\(\Rightarrow2.x=159\)
\(\Rightarrow x=159:2=\frac{159}{2}\)
a=bao nhiêu bạn
xin lỗi vì thếu đề
a bài 1 là bằng 3/11 nha