Viết các biểu thức sau dưới dạng lũy thừa
a/ 9 x 32 x \(\frac{1}{81}\)x 27
b/ 4 x 32: (23 x \(\frac{1}{16}\))
c/34 x 35 :\(\frac{1}{27}\)
d/ \(\frac{2^2.4.32}{\left(-2\right)^2.2^5}\)
Viết các biểu thức sau dưới dạng một lũy thừa:
6, 3/2 x 9/4 x 81/16
7, (1/2)^7 x 8 x 32 x 2^8
8, (-1/7)^4 x 125 x 5
9, 4 x 32 : (2^3 x 1/16)
10, (1/7)^2 x 1/7 x 49
6:=(3/2)*(3/2)^2*(3/2)^4=(3/2)^7
7: =(1/2)^7*2^3*2^5*2^8=2^9
8: =(-1/7)^4*5^4=(-5/7)^4
9: =2^2*2^5:(2^3/2^4)
=2^7/2=2^6
10: =(1/7)^3*7^2=1/7
Viết về dạng lũy thừa an
a, 2; 4; 16; 32; 128; 512
b, 9; 27; \(\frac{1}{81}\); 243
c, (4,25) : (23 \(\frac{1}{6}\))
d, \(\frac{\left(2^2\right)\cdot4\cdot32}{2^2\cdot25}\)
Viết các biểu thức sau dưới dạng một lũy thừa:
6; 3/2 x 9/4 x 81/16
7; (1/2)^7 x 8 x 32 x 2^8
8; (-1/7)^4 x 125 x 5
9; 4 x 32 : (2^3 x 1/16)
10; (1/7)^2 x 1/7 x 49
\(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)
\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8=\left(\dfrac{1}{7}\right)^7\times2^{16}\)
\(\left(-\dfrac{1}{7}\right)^4\times125\times5=\left(-\dfrac{1}{7}\right)^4\times5^3\times5=\left(-\dfrac{1}{7}\right)^4\times5^4\)
\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:2^3:2^{-4}=2^0\)
\(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^3\times7^3=1^3\)
6, \(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4\)
7,\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8\)
8,\(\left(-\dfrac{1}{7}\right) ^4\times125\times5=\left(\dfrac{1}{7}\right)^4\times5^3\times5\)
9,\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:\left[2^3\times\left(\dfrac{1}{2}\right)^4\right]\)
10, \(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times7^2\)
6) \(...=\dfrac{3}{2}.\left(\dfrac{3}{2}\right)^2.\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)
7) \(...=2^{-7}.2^3.2^5.2^8=2^9\)
8) \(...=\left(\dfrac{1}{7}\right)^4.5^3.5=\left(\dfrac{1}{7}\right)^4.5^4\)
9) \(...=2^2.2^5:\left(2^3.2^{-4}\right)=2^8\)
10) \(...=\left(\dfrac{1}{7}\right)^3.\left(\dfrac{1}{7}\right)^2=\left(\dfrac{1}{7}\right)^5\)
Viết biểu thức sau dưới dạng: (với a thuộc Q; m thuộc Z)
a, \(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
b,\(4^6.256^2.2^4\)
c,\(\left(\left(\frac{1}{9}:\frac{8}{27}\right):\frac{16}{48}\right):\frac{81}{128}\)
viết các biểu thức sau dưới dạng lũy thừa
a) 9 x 3 mũ 2 x 1/81 x 27
b) 4x32 :(2 mũ 3 x 1/16)
c)3 mũ 4 x 3 mũ 5 : 1/27
d) 81 mũ 11 x 3 mũ 17 / 27 mũ 10 x 9 mũ 15
Viết các biểu thức sau dưới dạng tổng hoặc hiệu các lập phương
\(\left(1-\frac{x}{5}\right)\left(\left(\frac{x}{5}\right)^2+\frac{x}{5}+1\right)\)
help!!!!!!!!!
\(\Leftrightarrow1-\frac{x^3}{125}\)
Nó là hằng đẳng thức mà .
Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
\(a,\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(b,\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
tìm x biết
a,\( |2-\frac{3}{2}x|-4=x+2\)
b,\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
c,\(2.3^x-405=3^{x-1}\)
d.\(\left(\frac{1}{81}\right)^x.27^{2.x}=\left(-9\right)^4\)
e,\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
g,\(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
h,\(|3x-5|=|\frac{1}{2}x+3|\)
i,\(\left(\frac{3}{4}-x\right)^3=\frac{1}{64}\)
a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)
=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)
=> \(\left|2-\frac{3}{2}x\right|=x+6\)
ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)
Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)
=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)
=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)
b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)
=> x = 1/4
hoặc x = 0 hoặc x = 1/2
viết các biểu thức sau dưới dạng lũy thừa: a) 9 x 3^2 x 1/81;b) 3^4 x 3^5 : 1/27; c) 2^6x4x32/(-2)^2x25
b: \(3^4\cdot3^5:\dfrac{1}{27}==3^9\cdot3^3=3^{12}\)