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16 tháng 7 2023

Câu 5:

\(D\left(2\right)=21a+9b-6a-4b\)

\(D\left(2\right)=\left(21a-6a\right)+\left(9b-4b\right)\)

\(D\left(2\right)=15a+5b\)

Mà: \(3a+b=18\Rightarrow b=18-3b\)

\(\Rightarrow D\left(2\right)=15a+5\left(18-3b\right)\)

\(D\left(2\right)=15a+90-15a\)

\(D\left(2\right)=90\)

Vậy: ...

16 tháng 7 2023

còn câu 3, với 4 ạ?

4:

D=6a+9b=3(2a+3b)=36

5: 

D=15a+5b=5(3a+b)=90

5 tháng 6 2018

Bài 1:

a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)

\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)

\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)

\(=\frac{-16}{7}\)

b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)

\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)

\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)

\(=\frac{10}{3}\)

5 tháng 6 2018

Bài 2:

a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)

b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)

\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)

c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)

\(=1-1\frac{14}{15}=\frac{14}{15}\)

Bài 3:

a) x/5 = 2/5

=> x =2

b) -4/x = 20/14 = 10/7

=> -4/x = 10/7

=> x.10 = (-4).7

x.10 = - 28

x= -28 :10

x= -2,8

c) 4/7 = 12/x = 12/ 21

=> 12/x = 12/21

=> x = 21

d) 3/7 = x / 21 = 9/21

=> x/21 = 9/21

=> x= 9

21 tháng 2 2018

1. a, => -12x+60+21-7x = 5

=> 81 - 19x = 5

=> 19x = 81 - 5 = 76

=> x = 76 : 19 = 4

Tk mk nha

17 tháng 2 2020

\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)

\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)

\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)

\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)

\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)

\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)

\(C=\frac{91}{9}\)

27 tháng 4 2020

\(A=-\frac{7}{21}+\left(1+\frac{1}{3}\right)\)

\(A=-\frac{1}{3}+1+\frac{1}{3}\)

\(A=\left(-\frac{1}{3}+\frac{1}{3}\right)+1=0+1=1\)

\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)

\(B=\frac{2}{15}+\frac{-1}{9}=\frac{6}{45}+\frac{-5}{45}=\frac{1}{45}\)

\(C=\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\left(-\frac{3}{5}\right)+\frac{2}{21}+\left(-\frac{10}{21}\right)+\frac{3}{20}\)

\(C=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\left(-\frac{3}{5}\right)+\frac{2}{21}+\left(-\frac{10}{21}\right)+\frac{3}{20}\)

\(C=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(C=0+0+\frac{3}{20}=\frac{3}{20}\)