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16 tháng 9 2018
a) Có \(\dfrac{2^8.9^2}{6^4.8^2}\)=\(\dfrac{2^8.3^4}{3^4.2^42^6}\)=\(\dfrac{2^8.3^4}{3^4.2^{10}}\)=\(\dfrac{2^8}{2^{10}}\)=\(\dfrac{1}{4}\)
b) Có \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right).2^3\)
= \(\dfrac{2^{-1}+3^{-1}}{2^{-1}-3^{-1}}+2^{-1}.2^3\)
=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{\dfrac{1}{2}-\dfrac{1}{3}}+\dfrac{1}{2}.2^3\)
= \(\dfrac{\dfrac{5}{6}}{\dfrac{1}{6}}+2^2\)=5+8=13
21 tháng 10 2016
a,15^8*9^7/27^7*25^4
=3^8*5^8*3^14/3^21*5^8
=3^22*5^8/3^21*5^8
=3
b,9^3/(3^4-3^3)^2
=3^6/3^8-3^6
=3^6/3^6*(3^2-1)
=3^6/3^6*8
=1/8
c,(1/2-2/3+3/4-2/5):x=11/30
=> 11/60:x=11/30
=> x=11/60:11/30
=> x=1/2
d,-3/4*x+0,7*x=1,25:1/8
=> x*(-3/4+0,7)=10
=> x*(-1/20)=10
=> x=10:(-1/20)=200
Mình làm tắt lắm mong bạn thông cảm.
14 tháng 7 2017
A) \(\dfrac{4^5.4^2}{16^4}=\dfrac{4^7}{\left(2^4\right)^4}=\dfrac{2^{14}}{2^{16}}=\dfrac{1}{4}\)
b)\(\dfrac{2^8.9^4}{6^6.8^3}=\dfrac{2^8.\left(3^2\right)^4}{2^6.3^6.\left(2^3\right)^3}=\dfrac{2^8.3^8}{2^{15}.3^6}=\dfrac{9}{128}\)
14 tháng 7 2017
c) \(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\dfrac{2^3.3^3+3^3.2^2+3^3}{-13}=\dfrac{3^3.\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3.13}{-13}=-9\)
25 tháng 7 2020
Câu a) số lớn lắm
b) \(3^{-3}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^x=3^3\)
=> \(3^x=3^3:\frac{1}{27}=3^3:\left(\frac{1}{3}\right)^3=3^3:\frac{1^3}{3^3}=3^3\cdot3^3=3^6\)
=> x = 6
b) \(\left(7x+2\right)^{-1}=3^{-2}\)
=> \(\frac{1}{7x+2}=\frac{1}{9}\)
=> 7x + 2 = 9
=> 7x = 7
=> x = 1
Bài 2:
a) \(3^4\cdot\frac{1}{729}\cdot81^3\cdot\frac{1}{9^2}\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot\left(3^4\right)^3\cdot\left(\frac{1}{3}\right)^4\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot3^{12}\cdot\left(\frac{1}{3}\right)^4=3^{16}\cdot\left(\frac{1}{3}\right)^{10}=\frac{3^{16}}{3^{10}}=3^6\)
b) \(\left(8\cdot2^5\right):\left(2^4\cdot\frac{1}{32}\right)=\left(2^3\cdot2^5\right):\left(2^4\cdot\left(\frac{1}{2}\right)^5\right)\)
\(=2^8:\left(2^4\cdot\frac{1^5}{2^5}\right)=2^8:\left(\frac{2^4}{2^5}\right)=2^8:2^{-1}=512\)
c) \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
d) Tương tự
25 tháng 5 2022
a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\)
\(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\)
b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\)
\(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\)
c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\)
\(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\)
e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\)
22 tháng 7 2017
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
23 tháng 7 2017
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
CW
3 tháng 8 2016
a) \(VT=12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
\(VP=18^{16}=2^{16}\cdot3^{32}\)
=> VT=VP
b) \(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
(đề sai)
c) \(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
\(VT=\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left[3^3\left(3-1\right)\right]^2}=\frac{1}{2^2}=\frac{1}{4}=VP\)
nhấn máy tính là ra