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YN
17 tháng 9 2021
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{5}-\frac{1}{16}\right)]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-[2.\frac{3}{16}]=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=2008\)
YN
17 tháng 9 2021
\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Rightarrow x=15\)
PT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
8 tháng 12 2018
Đặt biểu thức trong ngoặc là A
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{7.8.9.10}.\)
\(3A=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+\frac{6-3}{3.4.5.6}+...+\frac{10-7}{7.8.9.10}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{7.8.9}-\frac{1}{8.9.10}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{8.9.10}\Rightarrow A=\frac{1}{1.2.3.3}-\frac{1}{3.8.9.10}\)
Từ đó tính ra x . Bạn tự làm nốt nhé. Ngại tính
TT
2
MN
8 tháng 6 2021
\(\dfrac{1}{120}\cdot120+x:\dfrac{1}{3}=-4\)
\(\Leftrightarrow1+x\cdot3=-4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
10 tháng 6 2021
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
30 tháng 7 2018
\(a) \frac{4}{5}.x=\frac{8}{35}\)
\(\implies x= \frac{8}{35}:\frac{4}{5}\)
\(\implies x=\frac{8}{35}.\frac{5}{4}\)
\(\implies x=\frac{2}{7}\). Vậy \(x=\frac{2}{7}\)
\(b) \frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(\implies \frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(\implies \frac{3}{5}x=\frac{9}{14}\)
\(\implies x=\frac{9}{14}:\frac{3}{5}\)
\(\implies x=\frac{9}{14}.\frac{5}{3} \)
\(\implies x=\frac{15}{14}\). Vậy \(x=\frac{15}{14}\)
\(c) x-25\% x=0,5\)
\(\implies 75\% x=0,5\)
\(\implies \frac{3}{4}x=\frac{1}{2}\)
\(\implies x=\frac{1}{2}:\frac{3}{4}\)
\(\implies x=\frac{1}{2}.\frac{4}{3}\)
\(\implies x=\frac{2}{3}\). Vậy \(x=\frac{2}{3}\)
\(d) (\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}).120+x:\frac{1}{3}=-4\)
Có: \(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Thay vào ta có: \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(\implies 1+x:\frac{1}{3}=-4\)
\(\implies x:\frac{1}{3}=-5\)
\(\implies x=-5.\frac{1}{3}\)
\(\implies x=\frac{-5}{3}\). Vậy \(x=\frac{-5}{3}\)
~ Hok tốt a~
30 tháng 7 2018
\(\frac{4}{5}.x=\frac{8}{35}\) \(\frac{3}{5}x-\frac{1}{2}=\frac{1}{7}\)
\(x=\frac{8}{35}:\frac{4}{5}\) \(\frac{3}{5}x=\frac{1}{7}+\frac{1}{2}\)
\(x=\frac{8}{35}.\frac{5}{4}\) \(\frac{3}{5}x=\frac{9}{14}\)
\(x=\frac{2}{7}\) \(x=\frac{9}{14}:\frac{3}{5}\)
Vậy \(x=\frac{2}{7}\) \(x=\frac{9}{14}.\frac{5}{3}\)
\(x-25\%x=0.5\) \(x=\frac{15}{14}\)
\(x-\frac{1}{4}x=\frac{1}{2}\) Vậy \(x=\frac{15}{14}\)
\(x\left(1-\frac{1}{4}\right)=\frac{1}{2}\) \(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
\(x=\frac{1}{2}:\frac{3}{4}\) \(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\) \(x=\frac{2}{3}\) \(\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
Vậy \(x=\frac{2}{3}\) \(\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(1+x:\frac{1}{3}=-4\)
\(x:\frac{1}{3}=\left(-4\right)-1\)
\(x:\frac{1}{3}=-5\)
\(x=\left(-5\right).\frac{1}{3}\)
\(x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}\)
\(\left(12\cdot15-x\right)\cdot\dfrac{1}{4}=120\cdot\dfrac{1}{4}\)
\(\Rightarrow\left(180-x\right)\cdot\dfrac{1}{4}=30\)
\(\Rightarrow180-x=30:\dfrac{1}{4}\)
\(\Rightarrow180-x=120\)
\(\Rightarrow x=180-120\)
\(\Rightarrow x=60\)