Bài 31 Viết các số và dưới dạng các lũy thừa của cơ số 0,5

Lời giải:

Ta có:

Giải : Ta có:

(0,25)⁸ =[(0,5)²]⁸=(0,5)¹⁶

(0,125)⁴=[(0,5)³]⁴=(0,5)¹²

a) \({\left( {\frac{8}{9}} \right)^3} \cdot \frac{4}{3} \cdot \frac{2}{3} = {\left( {\frac{8}{9}} \right)^3}.\frac{8}{9} = {\left( {\frac{8}{9}} \right)^{3+1}}={\left( {\frac{8}{9}} \right)^4}\)

b) \({\left( {\frac{1}{4}} \right)^7} \cdot 0,25 = {\left( {0,25} \right)^7}.0,25 ={\left( {0,25} \right)^{7+1}}= {\left( {0,25} \right)^8}\)

c) \({( - 0,125)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^{6-1}}= {\left( {\frac{{ - 1}}{8}} \right)^5}\)

d) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)

\(\text{Xét công thức tổng quát }:x^4+\frac{1}{4}=\left(x^4+2.x^2.\frac{1}{2}+\frac{1}{4}\right)-x^2\)

\(=\left(x^2+\frac{1}{2}\right)^2-x^2=\left(x^2-x+\frac{1}{2}\right)\left(x^2+x+\frac{1}{2}\right)\)

Áp dụng vào B ta đc:

\(B=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(11^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(12^4+\frac{1}{4}\right)}\)

\(=\frac{\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)...\left(11^2-11+\frac{1}{2}\right)\left(11^2+11+\frac{1}{2}\right)}{\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)...\left(12^2-12+\frac{1}{2}\right)\left(12^2+12+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(110+\frac{1}{2}\right)\left(122+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(132+\frac{1}{2}\right)\left(156+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}\left(122+\frac{1}{2}\right)}{\left(132+\frac{1}{2}\right)\left(156+\frac{1}{2}\right)}=\frac{49}{16589}\)

ko biết có đúng ko!! hình như còn 1 cách là nhân 1 đa thức với 16 nữa thì phải lâu ko động đến bạn thử xem đc ko nhé

(0,25)⁸=(0,5²)⁸=(0,5)¹⁶

(0,125)⁴=(0,5³)⁴=(0,5)¹²

a: \(\frac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)

\(=\frac{2^{15}\cdot5^8-2^5\cdot5^9\cdot2^9}{2^{16}\cdot5^7+\left(2^2\cdot5\right)^8}=\frac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}\)

\(=\frac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\left(1+5\right)}=\frac{5}{2^2}\cdot\frac{-3}{6}=\frac54\cdot\frac{-1}{2}=-\frac58\)

b: \(\frac{\left(-0,25\right)^{-5}\cdot9^4\cdot\left(-2\right)^{-3}-2^{-2}\cdot6^9}{2^9\cdot3^6+6^6\cdot40}\)

\(=\frac{-2^{10}\cdot3^8\cdot\left(-\frac18\right)-\frac14\cdot2^9\cdot3^9}{2^9\cdot3^6+2^6\cdot3^6\cdot2^3\cdot5}=\frac{-2^7\cdot3^8-2^7\cdot3^9}{2^9\cdot3^6+2^9\cdot3^6\cdot5}\)

\(=\frac{-2^7\cdot3^8\left(1+3\right)}{2^9\cdot3^6\left(1+5\right)}=-\frac{1}{4\cdot3^2}\cdot\frac46=-\frac{1}{9\cdot6}=-\frac{1}{54}\)

Ta có:

\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)

=\(16.\left(\dfrac{5}{8}-0,25\right):\left(3\dfrac{1}{3}-2\dfrac{1}{4}\right)\)

=\(16.\left(\dfrac{5}{8}-\dfrac{1}{4}\right):\left(\dfrac{10}{3}-\dfrac{9}{4}\right)\)

=\(16.\dfrac{3}{8}:\dfrac{13}{12}\)

=\(6:\dfrac{13}{12}\)

=\(\dfrac{72}{13}\)

= 16 . (5/8-1/4) : (10/3-9/4)

= 16. 3/8:13/12

= 6:13/12

= 72/13

bài lớp mấy vậy nhìn nó khó khó