\(\frac{6}{8}=\frac{15}{x}\)
Bài 1:Tìm x,biết
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
b)\(x+\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}=\frac{2}{3}\)
c)\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}=\frac{5}{24}\)
d)\(8-\frac{8-\frac{8}{5}+\frac{8}{25}-\frac{8}{125}}{9-\frac{9}{5}+\frac{9}{25}-\frac{9}{125}}:\frac{161616}{151515}=\frac{4+\frac{4}{73}-\frac{4}{115}}{5+\frac{5}{73}-\frac{1}{23}}\)
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)
a)\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
b)\(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)
c)\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\)
d)\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
ai xong nhanh nhất và đúng em xin gửi 2 SP ạ
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
1. thực hiện phép tính
a) \(\frac{-1}{2}-\frac{9}{8}:0,75\) b)\(\frac{4}{3}.\frac{11}{15}+\frac{4}{15}.\frac{4}{3}-2\frac{1}{3}\) c)\(\frac{6}{8}+\frac{6}{56}+\frac{6}{140}+.....+\frac{6}{416}\)
2. tìm x
a) x ∈ Z, x∈ Ư (4) và x ∠1 b)x - \(\frac{2}{3}\)=20 % c) \((2\frac{1}{6}-x):\frac{2}{3}=\left(-\frac{1}{2}\right)^2\)
Giải pt : \(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\)
\(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\)
\(\Leftrightarrow3\left(5x-1\right)+5\left(2x+3\right)=2\left(x-8\right)-x\)
\(\Leftrightarrow15x-3+10x+15=2x-16-x\)
\(\Leftrightarrow15x+10x-2x+x=-16+3-15\)
\(\Leftrightarrow24x=-28\)
\(\Leftrightarrow x=\frac{-28}{24}=\frac{-7}{6}\)
Vậy ...
tìm x biết
1, \(x\cdot\frac{6}{25}=\frac{15}{-13}\)
2, \(x\div\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\)
3, \(\frac{3}{8}-\frac{1}{6}\cdot x=\frac{1}{4}\)
4, \(\frac{1}{3}+\frac{1}{2}\div x=-4\)
5, \(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(x\times\frac{6}{25}=\frac{15}{-13}\)
x=\(\frac{15}{-13}\div\frac{6}{25}\)
x=\(-\frac{125}{26}\)
các câu còn lại làm tương tự nha!!!
\(1.x.\frac{6}{25}=\frac{15}{-13}\\ x=\frac{15}{-13}:\frac{6}{25}\\ x=-\frac{125}{26}\)
\(2.x:\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\\ x:\frac{4}{10}=\frac{11}{45}\\ x=\frac{11}{45}.\frac{4}{10}\\ x=\frac{22}{225}\)
\(3.\frac{3}{8}-\frac{1}{6}.x=\frac{1}{4}\\ \frac{1}{6}.x=\frac{3}{8}-\frac{1}{4}\\ \frac{1}{6}.x=\frac{1}{8}\\ x=\frac{1}{8}:\frac{1}{6}\\ x=\frac{3}{4}\)
\(4.\frac{1}{3}+\frac{1}{2}:x=-4\\ \frac{1}{2}:x=-4-\frac{1}{3}=-\frac{13}{3}\\ x=\frac{1}{2}:\left(-\frac{13}{3}\right)=-\frac{3}{26}\)
\(5.x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}=\frac{5}{6}\\ x=\frac{5}{6}-\frac{7}{12}\\ x=\frac{1}{4}\)
tìm x biết
1, \(x\cdot\frac{6}{25}=\frac{15}{-13}\)
2, \(x\div\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\)
3, \(\frac{3}{8}-\frac{1}{6}\cdot x=\frac{1}{4}\)
4, \(\frac{1}{3}+\frac{1}{2}\div x=-4\)
5, \(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
giải phương trình sau:
a, \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
b,
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)
=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)
=> \(x-2004=0\)
=> \(x=2004\)
Vậy phương trình có nghiệm là x = 2004 .
b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)
=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
=> \(x-100=0\)
=> \(x=100\)
Vậy phương trình có nghiệm là x = 100 .
Tìm x biết
\(\frac{15}{x^6}:\frac{1}{8}=2,5:\frac{2}{5}x\)
4\(\frac{1}{2}\)-\(\frac{15}{10.9}-\frac{15}{9.8}-...-\frac{15}{3.2}-\frac{15}{2.1}\)
\(\frac{6}{2.5}+\frac{8}{5.9}+\frac{12}{9.15}\)
\(1-\frac{2}{2.1}-\frac{6}{2.5}-\frac{8}{5.9}-\frac{8}{9.13}\)