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Quỳnh Thơ
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Incursion_03
13 tháng 1 2019 lúc 21:33

\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{13.15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\left(1-\frac{1}{15}\right)\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\frac{14}{15}\left(x-1\right)=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow\frac{14}{15}x-\frac{14}{15}=\frac{6}{5}x-\frac{14}{15}\)

\(\Leftrightarrow-\frac{4}{15}x=\frac{28}{15}\)

\(\Leftrightarrow x=7\)

Thảo Karry
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công chúa bong bóng
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Nguyen Bao Linh
1 tháng 3 2017 lúc 16:40

a) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{15}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}x-\frac{7}{15}=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}x-\frac{7}{15}-\frac{3}{5}x+\frac{7}{15}=0\)

\(\Leftrightarrow\frac{8}{15}x=0\)

\(\Leftrightarrow x=0\)

Ngọc ANh
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Duong Thi Nhuong
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Chí Cường
7 tháng 1 2017 lúc 11:48

$\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{103.105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{103}-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{1}{2}.\left(1-\frac{1}{105}\right).\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}.\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow \frac{52}{105}x-\frac{52}{105}=\frac{3}{5}x-\frac{7}{15}\\ \Leftrightarrow x=-\frac{3}{11}$

Ngáo Nu
8 tháng 1 2017 lúc 12:11

b) Đặt \(A=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\)

A\(=\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{1}{3.103}+...+\frac{100}{10.110}\right)\)

A\(=\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)

A\(=\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)Đặt \(B=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{10}{100.110}\)

\(B=\frac{1}{10}\left(\frac{10}{1.11}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)

\(B=\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(B=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right]\)\(=\frac{1}{10}\left[\left(1+\frac{1}{2}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)\(B=10A\)

\(A.x=10A\)

\(=>x=10\)

kaito duy nguyen
9 tháng 1 2017 lúc 20:13

thanh niên ngu méo làm được bài này

Cao Tran Tieu Doan
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Pham Van Hung
3 tháng 3 2019 lúc 16:49

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)=\frac{7}{15}\)

Ta có: \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\Leftrightarrow\frac{7}{15}x=\frac{3}{5}x\)

\(\Leftrightarrow\frac{2}{15}x=0\Leftrightarrow x=0\)

Tập nghiệm: \(S=\left\{0\right\}\)

Giga Wizz
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Bùi Nhật Anh
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Nguyễn Hải Dương Anh
16 tháng 9 2020 lúc 13:53

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{7}\)

\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{7}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow\frac{1}{x+2}=-\frac{9}{7}\)

\(\Rightarrow-9\left(x+2\right)=7\)

\(\Rightarrow x+2=-\frac{7}{9}\)

\(\Rightarrow x=-\frac{25}{9}\)

Vậy \(x=-\frac{25}{9}\)

Khách vãng lai đã xóa
Nguyễn Ngọc Phương Linh
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BAN is VBN
25 tháng 3 2016 lúc 16:43

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{5}{11}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\Rightarrow1-\frac{1}{x+2}=\frac{5}{11}\div\frac{1}{2}=\frac{10}{11}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{10}{11}=\frac{1}{11}\Rightarrow x+2=11\Rightarrow x=11-2=9\)

Đợi anh khô nước mắt
25 tháng 3 2016 lúc 16:41

\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{x+\left(x+2\right)}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+2}\)

\(=1-\frac{1}{x+2}=\frac{5}{11}\)

\(\frac{1}{x+2}=1-\frac{5}{11}=\frac{6}{11}\)

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