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ND
Nguyễn Đăng Nhân
CTVHS
22 tháng 2 2023
a) \(\dfrac{\left(-3\right)^7\cdot2^8}{6^7}\)
\(=\dfrac{-1\cdot3^7\cdot2^8}{\left(2\cdot3\right)^7}=\dfrac{-1\cdot3^7\cdot2^7\cdot2}{2^7\cdot3^7}=-1\cdot2=-2\)
b) \(\dfrac{-3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}\)
\(=\dfrac{-3\cdot7\cdot7^3+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\left(-3\cdot7+1\right)}{7^3\left(7^2\cdot6-2\right)}=\dfrac{-3\cdot7+1}{7^2\cdot6-2}\)
\(=\dfrac{-21+1}{294-2}=\dfrac{-20}{290}=\dfrac{-2}{29}\)
b) \(\dfrac{5^3\cdot3^5}{5^3\cdot0,5+125\cdot2\cdot5}\)
\(=\dfrac{5^3\cdot3^5}{5^3\cdot0,5+5^3\cdot2\cdot5}=\dfrac{5^3\cdot3^5}{5^3\left(0,5+2\cdot5\right)}\)
\(=\dfrac{3^5}{0,5+2\cdot5}=\dfrac{243}{10,5}=\dfrac{162}{7}\)
NT
1
17 tháng 7 2018
Ta có :
\(\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^6+3.7^{15}}\)
\(=\dfrac{5.7^6\left(3.7^9-19.7^8\right)}{7^6\left(1+3.7^9\right)}\)
\(=\dfrac{5.7^8\left(3.7-19\right)}{1+3.7^9}\)
\(=\dfrac{5.7^8.2}{1+3.7^9}\)
\(=\dfrac{10.7^8}{1+3.7^8.7}\)
\(=\dfrac{10.7^8}{1+7^8.21}\)
NM
21 tháng 9 2023
\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)
\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)
\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)
13 tháng 6 2023
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
H9
HT.Phong (9A5)
CTVHS
13 tháng 10 2023
\(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5-\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
\(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=\left(3-5-6\right)-\left(\dfrac{1}{4}-\dfrac{7}{4}\right)-\dfrac{3}{2}+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\dfrac{6}{5}\)
\(=-8-\dfrac{6}{4}-\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{6}{5}\)
\(=-8-\dfrac{3}{2}-\dfrac{3}{2}+1+\dfrac{6}{5}\)
\(=-7-3+\dfrac{6}{5}\)
\(=\dfrac{6}{5}-10\)
\(=-\dfrac{44}{5}\)
9 tháng 1
a: \(\dfrac{2}{5}+\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{-2}{3}+\dfrac{1}{2}\)
\(=\dfrac{2}{5}-\dfrac{2}{5}+\dfrac{1}{2}=\dfrac{1}{2}\)
b: \(2,5-\left(-\dfrac{5}{6}\right)^0+\left(-\dfrac{1}{6}\right)^2\cdot\left(-3\right)\)
\(=\dfrac{5}{2}-1+\dfrac{1}{36}\cdot\left(-3\right)\)
\(=\dfrac{3}{2}-\dfrac{1}{12}=\dfrac{18}{12}-\dfrac{1}{12}=\dfrac{17}{12}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 10 2023
a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 10 2023
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55
28 tháng 7 2023
a: \(=6+\dfrac{4}{5}-1-\dfrac{2}{3}-3-\dfrac{4}{5}\)
\(=2-\dfrac{2}{3}=\dfrac{4}{3}\)
b: \(=7+\dfrac{5}{9}-2-\dfrac{3}{4}-3-\dfrac{5}{9}=2-\dfrac{3}{4}=\dfrac{5}{4}\)
c: =6+7/7-1-3/4-2-5/7
=3+2/7-3/4
=84/28+8/28-21/28
=84/28-13/28
=71/28
\(\dfrac{6^3.7-6^{13}.12}{6^3\left(5^3-5^2\right)}=\dfrac{6^3\left(7-12\right)}{5^3\left(125-25\right)}=\dfrac{-5}{100}=-\dfrac{1}{20}\)
\(=\dfrac{6^3\left(7-12\right)}{6^3\left(125-25\right)}=\dfrac{-5}{100}=-\dfrac{1}{20}\)