Question

Bài 2: Thực hiện phép tính (tính hợp lí nếu có)

a) \(\left(-2\right)^3+\frac{1}{2}\div\frac{1}{8}-\sqrt{25}+\left|-64\right|\)

b) \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)

c)\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

Giúp mk với!!! Mk cần gấp lắm!!!

Answer 1

a) Ta có: \(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)

\(=-8+\frac{1}{2}\cdot8-5+64\)

\(=-8+4-5+64=55\)

b) Ta có: \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)

\(=\left(\frac{-3}{4}+\frac{2}{7}\right)\cdot\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)

\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)

\(=0\cdot\frac{3}{2}=0\)

c) Ta có: \(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\frac{2^{10}\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}=\frac{2\left(2^9\cdot9^4-6^9\right)}{6^8\left(2^2+20\right)}=\frac{-1}{3}\)

Answer 2

a) ( -2 )³ + \(\frac{1}{2}:\frac{1}{8}\) - √25 + \(|-64|\)

= \(\frac{-8}{1}\) + \(\frac{1}{2}.\frac{8}{1}\) - \(\frac{5}{1}\) + \(\frac{64}{1}\)

= \(\frac{-16}{2}+\frac{1}{2}.\frac{8}{1}-\frac{10}{2}+\frac{128}{2}\)

= \(\frac{-16}{2}+\frac{8}{2}-\frac{10}{2}+\frac{128}{2}\)

= \(\frac{-16+8-10+128}{2}\) = \(\frac{110}{2}\) = 55

Answer 3

a/ = (-8) + \(\frac{1}{2}\) * 8 - 5 + 64

=(-8) + 4 - 5 + 64 = 55

b/ = ( \(\frac{-3}{4}\frac{ }{ }\) + \(\frac{2}{7}\) ) * 3 + ( \(\frac{-1}{4}+\frac{5}{7}\)) * 3

= 3*( \(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\) )

= 3*[\(\frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}\)]

= 3* (-1) + 1 = -3 +1 = -2

c/ = \(\frac{\text{(2^2 )^5 (3^2)^4 - 2 * (2*3)^9}}{2^{10}\cdot3^8+\left(2.3\right)^8\cdot2^2\cdot5}\) =\(\frac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\frac{2\cdot2\cdot3}{1\cdot1\cdot2^2\cdot5}=\frac{12}{20}=\frac{3}{5}\)

Answer 4

$b)\; \backslash (\backslash left(\backslash frac\{-3\}\{4\}+\backslash frac\{2\}\{7\}\backslash right):\backslash frac\{2\}\{3\}+\backslash left(\backslash frac\{-1\}\{4\}+\backslash frac\{5\}\{7\}\backslash right):\backslash frac\{2\}\{3\}\backslash )$

= \(\left(\frac{-3}{4}+\frac{2}{7}\right).\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right).\frac{3}{2}\)

= \(\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right).\frac{3}{2}\)

= \(\left[\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{2}{7}+\frac{5}{7}\right)\right].\frac{3}{2}\)

= \(\left[-1+1\right].\frac{3}{2}\)

= \(0.\frac{3}{2}\) = 0

Answer 5

sorry bn, câu b mk làm hơi bị lỗi chút, cái dòng đầu tiên là cái đề nha bn, còn phần còn lại bn trình bày theo cột dọc là đc ( bn thông cảm do mk viết nó bị lỗi nên nó dồn dô một dòng ), sorry nha !!!!!

c)\(\frac{4^6.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

= \(\frac{2^{12}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

= \(\frac{2^{12}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

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

= \(\frac{2^2-3}{1+5}\)

= \(\frac{1}{6}\)

Bài này khó nên chỗ nào bn ko hiểu thì viết dưới bình luận mk sẽ giải thích cho nha !!!!!

CHÚC BAN HỌC TỐT !!!!! (^_^)