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NX
16 tháng 8 2023
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
NX
16 tháng 8 2023
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
TD
2
18 tháng 3 2022
\(\frac{5}{3}-\frac{1}{4}+\frac{1}{3}-\frac{3}{4}\)
\(=\frac{5}{3}+\frac{-1}{4}+\frac{1}{3}+\frac{-3}{4}\)
\(=\left(\frac{5}{3}+\frac{1}{3}\right)+\left(\frac{-1}{4}+\frac{-3}{4}\right)\)
\(=\frac{6}{3}+\frac{-4}{4}\)
\(=2+\left(-1\right)\)
\(=1\)
14 tháng 12 2021
a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3
b)
Nhân 4 vào hai vế ta được:
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3
=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]
=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]
=n.(n+1).(n+2)
=>S=[n.(n+1).(n+2)] /3
ND
Nguyễn Đức Trí
VIP
24 tháng 8 2023
a) \(...=-\dfrac{1}{4}.\dfrac{4}{17}.\left(-\dfrac{63}{21}\right).\left(-\dfrac{7}{12}\right)\)
\(=-\dfrac{1}{17}.\dfrac{63}{21}.\dfrac{7}{12}\)
\(=-\dfrac{7}{68}\)
b) \(...=-\dfrac{2}{5}.\dfrac{4}{15}-\dfrac{3}{10}.\dfrac{4}{15}\)
\(=\dfrac{4}{15}\left(-\dfrac{2}{5}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}\left(-\dfrac{4}{10}-\dfrac{3}{10}\right)\)
\(=\dfrac{4}{15}.\left(-\dfrac{7}{10}\right)=-\dfrac{14}{75}\)
c) \(...=21-\dfrac{15}{4}:\left(\dfrac{9}{24}-\dfrac{4}{24}\right)\)
\(=21-\dfrac{15}{4}:\dfrac{5}{24}\)
\(=21-\dfrac{15}{4}.\dfrac{24}{5}\)
\(=21-3.6=3\)
d) \(...=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right).\dfrac{7}{3}\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)
\(=\dfrac{7}{3}\left(-\dfrac{3}{4}-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right)\)
\(=\dfrac{7}{3}\left(-1+1\right)=0\)
26 tháng 6 2023
a: =2/3+1/5*10/7
=2/3+2/7
=14/21+6/21=20/21
b: \(=\dfrac{1}{2}\cdot\dfrac{-3+2}{4}=\dfrac{1}{2}\cdot\dfrac{-1}{4}=\dfrac{-1}{8}\)
c: \(=\dfrac{3}{4}+\dfrac{9}{5}:\dfrac{3}{2}-1\)
=-1/4+9/5*2/3
=-1/4+18/15
=-1/4+6/5
=-5/20+24/20=19/20
d: \(=\dfrac{3}{2}\cdot\left(\dfrac{7}{3}-\dfrac{5}{3}\cdot4\right)\)
\(=\dfrac{7}{2}-\dfrac{5}{2}\cdot4=\dfrac{7}{2}-\dfrac{20}{2}=\dfrac{-13}{2}\)
Sửa đề: 1/2*1/3+1/3*1/4+...+1/19*1/20
=1/2-1/3+1/3-1/4+...+1/19-1/20
=1/2-1/20=10/20-1/20=9/20