\(A=\)\(\frac{4^5.9^4-2.6^9}{2^...">
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18 tháng 6 2019

#)Giải :

a) \(A=\frac{4^5.9^4-2^6.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=-\frac{1}{3}\)

\(a,A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-1}{3}\)

Học tốt!!!!!!!!!!!!!

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

9 tháng 2

Câu 1:

A = (4/9 + 5/6) : (7/8 - 7/12)

A = (8/18 + 15/18) :(21/24 - 14/24)

A = 23/18 : 7/24

A = 23/18 x 24/7

A = 92/21

9 tháng 2

Câu B = 3/4.8/9.15/16...9999/10000

B = \(\frac{1.3}{2.2}\).\(\frac{2.4}{3.3}\)....\(\frac{99.101}{100.100}\)

B = \(\frac{1.2.3\ldots99}{2.3\ldots100}\). \(\frac{3.4.\ldots101}{2.3\ldots100}\)

B = \(\frac{1}{100}\) x \(\frac{101}{2}\)

B = \(\frac{101}{200}\)

24 tháng 6 2020

bạn tự làm đi tính toán thôi mà

16 tháng 8 2025

15 tháng 8 2025

15 tháng 8 2025

9 tháng 2

Câu a:

(4 5/37 - 3 4/5 + 8 15/29) - (3 5/37 - 6 14/29)

= (4 5/37 - 3 5/37) + (8 15/29 + 6 14/29) - 3 4/5

= 1 + 15 - 3 4/5

= 16 - 3 - 4/5

= 13 - 4/5

= 61/5

9 tháng 2

Câu b:

(2\(\frac56\) + 1\(\frac49\)) : (10\(\frac{1}{12}\) - 9\(\frac12\))

= (17/6 + 13/9) :(121/12 - 19/2)

= (51/18 + 26/18) :(121/12 - 114/2)

= 77/18 : 7/12

= 77/18 x 12/7

= 22/3

13 tháng 1

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13 tháng 1

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29 tháng 3 2025

a,\(\frac{31}{1000}\)

b,\(\frac{8}{108}\)

c,0

a,\(\frac{49}{97}\)

b,\(\frac{-1}{4751}\)

S
9 tháng 3 2025

\(a;\dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{2}{3}:\dfrac{3}{2}\\ \dfrac{3}{2}x-\dfrac{2}{3}=\dfrac{4}{9}\\ \dfrac{3}{2}x=\dfrac{4}{9}+\dfrac{2}{3}=\dfrac{10}{9}\\ x=\dfrac{10}{9}:\dfrac{3}{2}=\dfrac{20}{27}\\ b;\left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=1-\dfrac{4}{5}\\ \left(\dfrac{9}{11}-x\right):\left(-\dfrac{10}{11}\right)=\dfrac{1}{5}\\ \dfrac{9}{11}-x=\dfrac{1}{5}\cdot\left(-\dfrac{10}{11}\right)\\ \dfrac{9}{11}-x=-\dfrac{2}{11}\\ x=\dfrac{9}{11}-\left(-\dfrac{2}{11}\right)=\dfrac{9}{11}+\dfrac{2}{11}\\ x=1\\ c;-\dfrac{11}{12}x+\dfrac{3}{4}=-\dfrac{1}{6}\\ -\dfrac{11}{12}x=-\dfrac{1}{6}-\dfrac{3}{4}\\ -\dfrac{11}{12}x=-\dfrac{11}{12}\\ x=\left(-\dfrac{11}{12}\right):\left(-\dfrac{11}{12}\right)=1\)

\(d;-\dfrac{5}{4}-\left(1\dfrac{1}{2}+x\right)=4,5\\ \Leftrightarrow-\dfrac{5}{4}-\left(\dfrac{3}{2}+x\right)=4,5\\\dfrac{3}{2}+x=-\dfrac{5}{4}-4,5\\ \dfrac{3}{2}+x=-\dfrac{23}{4}\\ x=-\dfrac{23}{4}-\dfrac{3}{2}\\ x=-\dfrac{29}{4}\\ đ;\left(\dfrac{3}{4}-x:\dfrac{2}{15}\right)\cdot\dfrac{1}{5}=-2,6\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-2,6:\dfrac{1}{5}\\ \dfrac{3}{4}-x:\dfrac{2}{15}=-13\\ x:\dfrac{2}{15}=\dfrac{3}{4}-\left(-13\right)\\ x:\dfrac{2}{15}=\dfrac{55}{4}\\ x=\dfrac{55}{4}\cdot\dfrac{2}{15}=\dfrac{11}{6}\\ e;3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}=-\dfrac{10}{3}\)

\(f;\left(1-2x\right)\cdot\dfrac{4}{5}=\left(-2\right)^3\\ \left(1-2x\right)\cdot\dfrac{4}{5}=-8\\ 1-2x=-8:\dfrac{4}{5}=-10\\ 2x=1-\left(-10\right)=11\\ x=\dfrac{11}{2}\\ g;\dfrac{1}{6}-\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{6}-\dfrac{1}{8}=\dfrac{1}{24}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{24}\Rightarrow x=\dfrac{3}{4}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{1}{24}\Rightarrow x=\dfrac{7}{12}\end{matrix}\right.\)