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![](https://rs.olm.vn/images/avt/0.png?1311)
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
c) x=-2 nha
d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)
=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mình biến đởi phần trong |......| rồi bạn thay vào nha
1/30 + 1/42 + 1/56 + 1/72 +1/ 90 + 1/110 + 1/132
=1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 +1/ 10.11
=1/5 -1/6 +1/6 - 1/7 +......+1/10 - 1/11
=1/5 - 1/11=11/55 - 5/55 =6/ 55
thay vào |....|=> |6/55 - x | = 2/3 => mở ra 2 trường hợp mà tính nha
chúc hok tốt
=>(1/5.6+1/6.7+1/7.8+1/9.10+1/10.11+1/11.12)-x=2/3
=>(1/5-1/+1/6-1/7+...+1/11-1/12)-x=2/3
=>(1/5-1/12)-x=2/3
=>7/60-x=2/3
=>x=7/60-2/3
=>x=-11/20
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (9-8x)x2=100
9-8x =100:2
9-8x =50
8x =9-50
8x =-41
x =-41:8
x = \(\frac{-41}{8}\)
b, (20-x)x2=60
20-x =60:2
20-x =30
x =20-30
x =-10
c, \(\frac{1}{2}x+\frac{8}{4}=\frac{100}{25}\)
\(\frac{1}{2}x+2=4\)
\(\frac{1}{2}x=4-2\)
\(\frac{1}{2}x=2\)
x =2:\(\frac{1}{2}\)
x =4
d, (90-x)10=1000
90-x =1000:10
90-x =100
x =90-100
x=-10
****
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{5.6}+\frac{3}{6.7}+......+\frac{3}{11.12}=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{11.12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{11}-\frac{1}{12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\left(\frac{1}{5}-\frac{1}{12}\right)=\frac{1}{6}X\)
\(\Rightarrow3.\frac{7}{60}=\frac{1}{6}X\)
\(\Rightarrow\frac{21}{60}=\frac{1}{6}X\)
\(\Rightarrow X=\frac{21}{60}\div\frac{1}{6}=\frac{21}{10}\)
Vậy \(X=\frac{21}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Tìm x
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=x\)
\(\Rightarrow1-\frac{1}{100}=x\)
\(\Rightarrow x=\frac{99}{100}\)
\(2.Tính\)
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
học vui!!
Xin lỗi nha. Bài 1 mk làm sai. Lại nè:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)
\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)=x\)
\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=x\)
\(\frac{1}{3}.\left(1-\frac{1}{100}\right)=x\)
\(\frac{1}{3}\cdot\frac{99}{100}=x\)
\(\frac{33}{100}=x\)
\(M=\frac{9^2}{9.10}.\frac{10^2}{10.11}.\frac{11^2}{11.12}.....\frac{100^2}{100.101}\)
\(=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}.....\frac{100}{101}=\frac{9}{101}\)