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Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
a, \(\dfrac{38}{7}\) + ( \(\dfrac{16}{7}\) - \(\dfrac{5}{3}\))
= \(\dfrac{38}{7}\) + \(\dfrac{16}{7}\) - \(\dfrac{5}{3}\)
= \(\dfrac{54}{7}\) - \(\dfrac{5}{3}\)
= \(\dfrac{162}{21}\) - \(\dfrac{35}{21}\)
= \(\dfrac{127}{21}\)
b, \(\dfrac{29}{9}\) - ( \(\dfrac{14}{9}\) + \(\dfrac{11}{9}\) )
= \(\dfrac{29-14-11}{9}\)
= \(\dfrac{4}{9}\)
c, ( \(\dfrac{2}{7}\) + \(\dfrac{3}{8}\) + \(\dfrac{12}{7}\) + \(\dfrac{5}{8}\) ) - \(\dfrac{19}{63}\)
= ( \(\dfrac{2+12}{7}\) + \(\dfrac{3+5}{8}\)) - \(\dfrac{19}{63}\)
= \(\left(1+2\right)\) - \(\dfrac{19}{63}\)
= \(3-\dfrac{19}{63}\)
= \(\dfrac{189}{63}\) - \(\dfrac{19}{63}\)
= \(\dfrac{170}{63}\)
d, \(\dfrac{3}{7}\) . \(\dfrac{4}{9}\) + \(\dfrac{4}{9}\) . \(\dfrac{4}{7}\) - \(\dfrac{4}{9}\)
= \(\dfrac{4}{9}\) . ( \(\dfrac{3}{7}\) + \(\dfrac{3}{7}\) ) - \(\dfrac{4}{9}\)
= \(\dfrac{4}{9}\) . \(1-\dfrac{4}{9}\)
= \(\dfrac{4}{9}\) - \(\dfrac{4}{9}\)
= 0
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
\(\frac{9}{8}-\frac{-11}{8}=\frac{9-\left(-11\right)}{8}=\frac{9+11}{8}=\frac{20}{8}=\frac{5}{2}\)
\(-\frac{1}{2}-\frac{5}{2}=\frac{-1-5}{2}=\frac{-6}{2}=-3\)
\(\frac{4}{3}-\frac{7}{3}=\frac{4-7}{3}=\frac{11}{3}\)
\(\frac{3}{-5}-\frac{-8}{10}=\frac{6}{-10}-\frac{-8}{10}=\frac{-6}{10}-\frac{-8}{10}=\frac{-6-\left(-8\right)}{10}=\frac{-6+8}{10}=\frac{2}{10}=\frac{1}{5}\)
\(\frac{5}{7}-\frac{-3}{21}=\frac{5}{7}-\frac{-1}{7}=\frac{5-\left(-1\right)}{7}=\frac{5+1}{7}=\frac{6}{7}\)
\(\frac{4}{-3}-\frac{9}{27}=\frac{4}{-3}-\frac{1}{3}=\frac{-4}{3}-\frac{1}{3}=\frac{-4-1}{3}=\frac{-5}{3}\)
\(\frac{7}{29}-\frac{9}{29}=\frac{7-9}{29}=\frac{2}{29}\)
\(\frac{-7}{22}-\frac{9}{22}=\frac{-7-9}{22}=\frac{-16}{22}=\frac{-8}{11}\)
\(\frac{-23}{7}-\frac{31}{7}=\frac{-23-31}{7}=\frac{-54}{7}\)
a/ x-2+7=-27
x-2=-34
x=-32
b/ Đề bị điên ak?
c/ -2x=2
x=-1
d/ ko hỉu đề
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)