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2: \(=\dfrac{59}{10}\cdot\dfrac{2}{3}-\left[\dfrac{7}{3}\cdot\left(4+\dfrac{1}{2}-2\right)\right]\cdot\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{4}{3}\cdot\dfrac{5}{2}=\dfrac{59}{15}-\dfrac{10}{3}=\dfrac{59-50}{15}=\dfrac{9}{15}=\dfrac{3}{5}\)

4: \(=254+254\cdot399+399\)

\(=254\cdot400+399\)

\(=101600+399=101999\)

5: \(=638+3900-41=3900+597=4497\)

2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)

3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)

=2/7-2/7=0

22 tháng 8 2019

1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)

  \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

 \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

 \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)

\(x=\frac{45}{4}+\frac{9}{4}\)

\(x=\frac{27}{2}\)

22 tháng 8 2019

Bước cưối 58/21 minh man viết nhầm nên sai 

Tính giá trị biểu thức :1. \(A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\) 2. \(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)3. \(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)4. \(D=(\frac {150}{1111}+\frac{5}{75}-\frac{14}{77})(\frac{1}{5}-\frac{1}{6}-\frac{1}{30})...
Đọc tiếp

Tính giá trị biểu thức :

1. \(A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\) 

2. \(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)

3. \(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

4. \(D=(\frac {150}{1111}+\frac{5}{75}-\frac{14}{77})(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}) \)

5. Cho \(M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right);N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\). Tính \(P=M-N\)

6. \(E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)

7. \(F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

8. \(G=\left[\frac{\left(6-4\frac{1}{2}\right):0,03}{\left(3\frac{1}{20}-2,65\right).4+\frac{2}{5}}-\frac{\left(0,3-\frac{3}{20}\right).1\frac{1}{2}}{\left(1,88+2\frac{3}{25}\right).\frac{1}{80}}\right]:\frac{49}{60}\)

9. \(H=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)

10. \(I=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\)

11. \(k=\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{999}\right)\)

12. \(L=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}...\)(98 thừa số)

13. \(M=-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{3}}}}\)

14. \(N=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)

15. \(P=\left(\frac{1}{4}-1\right)\left(\frac{1}{5}-1\right)...\left(\frac{1}{2000}-1\right)\left(\frac{1}{2001}-1\right)\)

16. \(Q=\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\right):\left(\frac{1}{1004.2006}+\frac{1}{1005.2005}+...+\frac{1}{2006.1004}\right)\)

3
2 tháng 5 2018

\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)

\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)

\(=\frac{2}{4}=\frac{1}{2}\)

\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)

\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)

\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)

\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)

\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)

\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)

\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)

\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)

\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\)               \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)

\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)                         \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)

\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\)                     \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)

\(=\frac{58}{7}-\frac{487}{63}\)                                          \(=\frac{577}{45}-\frac{280}{45}\)

\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\)                             \(=\frac{33}{5}\)

\(P=M-N\)

\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)

\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)

\(\Rightarrow P=\frac{-272}{45}\)

Vậy P = \(\frac{-272}{45}\)

\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)

\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)

\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)

\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)

\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)

\(=\frac{3}{8}+\frac{5}{8}=1\)

Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !

9 tháng 2 2018
sao không tự làm một số bài dễ đi
2 tháng 7 2019

a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)

\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)

\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)

\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)

\(=6:\left(-\frac{6}{12}\right)\)

\(=6:\left(-\frac{1}{2}\right)\)

\(=-12\)

b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )

=3/5: (-11/30) + 3/5 : (-7/5) 

=3/5:[-11/30+(-7/5)]

=3/5:53/30

=18/53

c) (1/2-13/14):5/7-(-2/21+1/7):5/7

= -3/7:5/7-1/21:5/7

=(-3/7-1/21):5/7

=-10/21:5/7

=-2/3

câu b vá c mình làm tắt nha. chúc bạn học tốt

26 tháng 6 2020

help me

26 tháng 6 2020

a)  45/343

b)  -1/4

c)  24/13

d)  -4 

bạn cho mình đi