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b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{3}{9}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\frac{2}{9}\)
\(x-2=\frac{16}{9}\cdot\frac{9}{2}\)
\(x-2=8\)
\(x=8+2\)
\(x=10\)
Vậy \(x=10\)
\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\)\(=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{2}{9}\right)=\frac{16}{9}\)
2(x-2)=16
x-2=8
x=10
\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x+2}{42} +\frac{x+2}{56}+\frac{x+2}{72}=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{9}\right)=\frac{16}{9}\)
\(\Rightarrow\left(x-2\right)\frac{4}{9}=\frac{16}{9}\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
Ai k mk mk k lại
\(\left(x-2\right):\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right):\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right):\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{32}{91}\)
\(x=\frac{32}{91}+2\)
\(x=\frac{212}{91}\)
1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
=1-1/9
=8/9
A) \(x = {7 \over 10}- {8 \over10} \)
\(x = {-1 \over 10}\)
B)\({2 \over3}x = 2{5 \over 6}-{3 \over4}\)
\({2 \over3}x = {25 \over 12}\)
\(x = {25 \over 12}/{2 \over3} \)
\(x = {25\over 8}\)
2/ Tính tổng:
\( = {8 \over 9}\)
\(\frac{20}{9}-x=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)
\(\Rightarrow\frac{20}{9}-x=\frac{2}{9}\Rightarrow x=\frac{20}{9}-\frac{2}{9}=\frac{18}{9}=2\)
Vậy x = 2.
k cho mk nha
\(2\frac{2}{9}-x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(2\frac{2}{9}-x=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(2\frac{2}{9}-x=\frac{1}{3}-\frac{1}{9}\)
\(2\frac{2}{9}-x=\frac{3}{9}-\frac{1}{9}\)
\(2\frac{2}{9}-x=\frac{2}{9}\)
\(x=2\frac{2}{9}-\frac{2}{9}\)
\(x=2\)
Vậy x = 2
a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)