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24 tháng 6

\(4\dfrac{7}{5741}\cdot\dfrac{1}{3759}-\dfrac{4}{3759}\cdot1\dfrac{2}{5741}+\dfrac{1}{3759}+\dfrac{1}{3759\cdot5741}\\ =\dfrac{22971}{5741}\cdot\dfrac{1}{3759}-\dfrac{1}{3759}\cdot\dfrac{22972}{5741}+\dfrac{1}{3759}\cdot\dfrac{5741}{5741}+\dfrac{1}{3759}\cdot\dfrac{1}{5741}\\ =\dfrac{1}{3759}\cdot\left(\dfrac{22971}{5741}-\dfrac{22972}{5741}+\dfrac{5741}{5741}+\dfrac{1}{5741}\right)\\ =\dfrac{1}{3759}\cdot\dfrac{5741}{5741}=\dfrac{1}{3759}\cdot1=\dfrac{1}{3759}\)

Đặt 3759=a; 5741=b

Theo đề, ta có: \(E=4\dfrac{7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\left(1+\dfrac{2}{b}\right)+\dfrac{1}{a}+\dfrac{1}{ab}\)

\(=\dfrac{4b+7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\dfrac{b+2}{b}+\dfrac{b+1}{ab}\)

\(=\dfrac{4b+7-4b-8+b+1}{ab}=\dfrac{b}{ab}=\dfrac{1}{a}=\dfrac{1}{3759}\)

20 tháng 5 2018

\(xy\left(x-y\right)+yz\left(y-z\right)+xz\left(z-x\right)\\ =xy\left(x-y\right)+yz\left[-\left(x-y\right)-\left(z-x\right)\right]+xz\left(z-x\right)\\ =xy\left(x-y\right)-yz\left(x-y\right)-yz\left(z-x\right)+xz\left(z-x\right)\\ =\left(x-y\right)\left(xy-yz\right)+\left(z-x\right)\left(xz-yz\right)\\ =y\left(x-y\right)\left(x-z\right)+z\left(z-x\right)\left(x-y\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)

16 tháng 7 2019

(4568 + 3759) - (4563 + 3759)

= 4568 + 3759 - 4563 - 3759

= 3

16 tháng 7 2019

Làm lại 

(4568 + 3759) - (4568 + 3759)

= 4568 + 3759 - 4563 + 3759

= 5

Tính giá trị biểu thức : 1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\) 2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\) 3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\) 4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\) 5. Cho...
Đọc tiếp

Tính giá trị biểu thức :

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

2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\)

4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\)

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

6. \(E=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)

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

8. \(G=\text{[}\dfrac{\left(6-4\dfrac{1}{2}\right):0,03}{\left(3\dfrac{1}{20}-2,65\right)\cdot4+\dfrac{2}{5}}-\dfrac{\left(0,3-\dfrac{3}{20}\right)\cdot1\dfrac{1}{2}}{\left(1,88+2\dfrac{3}{25}\right)\cdot\dfrac{1}{80}}\text{]}:\dfrac{49}{60}\)

9. \(H=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{4\cdot5\cdot6}+...+\dfrac{1}{98\cdot99\cdot100}\)

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

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

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

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

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

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

16. \(Q=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2005\cdot2006}\right):\left(\dfrac{1}{1004\cdot2006}+\dfrac{1}{1005\cdot2005}+...+\dfrac{1}{2006\cdot1004}\right)\)

2
27 tháng 11 2017

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

2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)

3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)

4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)

29 tháng 4 2022

hôi lì sít

22 tháng 6 2022

a) A=[27(14−13)]:[27(13−25)]=(14−13):(13−25)=114.
b) B=34(15−27−13+27)15(27+13)−13(27+13)=34(15−13)(15−13)(27+13)=11152.

13 tháng 7 2022

a) \mathrm{A}=\left[\dfrac{2}{7}\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right]:\left[\dfrac{2}{7}\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right]=\left(\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{1}{3}-\dfrac{2}{5}\right)=1 \dfrac{1}{4}.
b) \mathrm{B}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{3}+\dfrac{2}{7}\right)}{\dfrac{1}{5}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)-\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{3}\right)}{\left(\dfrac{1}{5}-\dfrac{1}{3}\right)\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=1 \dfrac{11}{52}

11 tháng 7 2017

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}......1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5.....9.11}\) ( Bước này bạn bỏ đi cũng được )

\(=\dfrac{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}\)

\(=\dfrac{\left(2.3.4.....9\right).10.2.\left(3.4.5.....10\right)}{1.\left(2.3.4.....9\right).\left(3.4.5.....10\right).11}\)

\(=\dfrac{10.2}{1.11}=\dfrac{20}{11}=1\dfrac{9}{11}\)

11 tháng 7 2017

\(1\dfrac{1}{3}.1\dfrac{1}{8}.1\dfrac{1}{15}.....1\dfrac{1}{99}\)

\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.....\dfrac{100}{99}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.....\dfrac{10.10}{9.11}\)

\(=\dfrac{2.2.3.3.4.4.....10.10}{1.3.2.4.3.5....9.11}\)

\(=\dfrac{2.3.4....10}{1.2.3....9}.\dfrac{2.3.4...10}{3.4.5....11}\)

\(=10.\dfrac{2}{11}=\dfrac{20}{11}\)