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giúp mik vs, mik bik các pạn giờ này đang ngủ rùi nhưng giúp mik lần này thui.yêu các pạn nhìu
\(5\frac{1}{2}+\left(-3\right)=\frac{11}{2}+\frac{-3}{1}\)\(=\frac{11}{2}+\frac{-6}{2}=\frac{5}{2}\)\(;\)
\(4\frac{9}{11}+\left(-2\frac{1}{11}\right)=\frac{53}{11}+\frac{-23}{11}\)\(=\frac{30}{11}\)\(;\)
\(2\frac{1}{2}+\left(-6\right)=\frac{5}{2}+\frac{-6}{1}\)\(=\frac{5}{2}+\frac{-12}{2}=\frac{-7}{2}\)\(;\)
\(\left(-\frac{4}{5}\right)+\frac{1}{2}=\frac{-4}{5}+\frac{1}{2}\)\(=\frac{-8}{10}+\frac{5}{10}=\frac{-3}{10}\)\(;\)
\(4,3-\left(-1,2\right)=4,3+1,2=5,5\)\(=\frac{55}{10}=\frac{11}{2}\)\(;\)
\(0-\left(-0,4\right)=0+0,4=0,4\)\(=\frac{4}{10}=\frac{2}{5}\)\(;\)
\(\frac{-2}{3}-\frac{-1}{3}=\frac{-2}{3}+\frac{1}{3}=\frac{-1}{3}\)\(;\)
\(\frac{-1}{2}-\frac{-1}{6}=\frac{-1}{2}+\frac{1}{6}\)\(=\frac{-3}{6}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}\)\(;\)
\(x+\frac{1}{3}=\frac{3}{4}\) \(;\) \(x-\frac{2}{5}=\frac{5}{7}\) \(;\)
\(x=\frac{3}{4}-\frac{1}{3}\) \(x=\frac{5}{7}+\frac{2}{5}\)
\(x=\frac{5}{12}\) \(x=\frac{39}{35}\)
\(-x-\frac{2}{3}=-\frac{6}{7}\) \(;\) \(\frac{4}{7}-x=\frac{1}{3}\)
\(\frac{6}{7}-\frac{2}{3}=x\) \(\frac{4}{7}-\frac{1}{3}=x\)
\(\frac{4}{21}=x\) \(\Leftrightarrow\)\(x=\frac{4}{21}\) \(\frac{5}{21}=x\)\(\Leftrightarrow\)\(x=\frac{5}{12}\)
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt
d) \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=-3+\frac{3}{2}+1-5+\frac{6}{5}-\frac{3}{2}\)
\(=-7+\frac{6}{5}=-\frac{29}{5}\)
a) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{3}.\frac{102}{103}\)
\(=\frac{34}{103}\)
b) \(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{2000.1999}-\left(\frac{1}{1999.1998}+\frac{1}{1998.1997}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)(*)
Đặt biểu thức trong ngoặc là A ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1997.1998}+\frac{1}{1998.1999}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1997}-\frac{1}{1998}+\frac{1}{1998}-\frac{1}{1999}\)
\(A=1-\frac{1}{1999}\)
\(A=\frac{1998}{1999}\)
Thay vào biểu thức (*) ta có :
\(\frac{1}{2000.1999}-\frac{1998}{1999}\)
\(=\frac{1}{3998000}-\frac{1998}{1999}\)
\(=\frac{-3995999}{3998000}\)
c) \(\frac{-1}{3}+\frac{-1}{15}+\frac{-1}{35}+\frac{-1}{63}+...+\frac{-1}{9999}\)
\(=\frac{-1}{1.3}+\frac{-1}{3.5}+\frac{-1}{5.7}+\frac{-1}{7.9}+...+\frac{-1}{99.101}\)
\(=\frac{-1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(=\frac{-1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{-1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{-1}{2}.\frac{100}{101}\)
\(=\frac{-50}{101}\)
_Chúc bạn học tốt_
B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7 + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15
= 0 -0-0-0-0+7/9 +13/15
= 74/45
Bài 1:
a) \(\left(\frac{1}{2}\right)^2\) và \(\left(\frac{1}{2}\right)^5\)
Ta có: \(\left(\frac{1}{2}\right)^2=\frac{1}{4}.\)
\(\left(\frac{1}{2}\right)^5=\frac{1}{32}.\)
Vì \(\frac{1}{4}< \frac{1}{32}.\)
=> \(\left(\frac{1}{2}\right)^2< \left(\frac{1}{2}\right)^5.\)
b) \(\left(2,4\right)^3\) và \(\left(2,4\right)^2\)
Ta có: \(\left(2,4\right)^3=13,824.\)
\(\left(2,4\right)^2=5,76.\)
Vì \(13,284>5,76.\)
=> \(\left(2,4\right)^3>\left(2,4\right)^2.\)
c) \(\left(-1\frac{1}{2}\right)^2\) và \(\left(-1\frac{1}{2}\right)^3\)
Ta có: \(\left(-1\frac{1}{2}\right)^2=\left(-\frac{3}{2}\right)^2=\frac{9}{4}.\)
\(\left(-1\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3=-\frac{27}{8}.\)
Vì số dương luôn lớn hơn số âm nên \(\frac{9}{4}>-\frac{27}{8}.\)
=> \(\left(-1\frac{1}{2}\right)^2>\left(-1\frac{1}{2}\right)^3.\)
Chúc bạn học tốt!