Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3\\ =\left[\left(\dfrac{2}{3}\right)^2\right]^4:\left(\dfrac{4}{9}\right)^3\\ =\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3\\ =\dfrac{4}{9}\)
\(27^3:3^2\\ =\left(3^3\right)^3:3^2\\ =3^9:3^2\\ =3^7\\ =2187\)
\(\left(-\dfrac{3}{5}\right)^4:\left(\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{3}{5}:\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{6}{25}\right)^4\)
\(\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{9}{25}\right)^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left[\left(\dfrac{3}{5}\right)^2\right]^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{3}{5}\right)^{10}\\ =\left(\dfrac{3}{5}\right)^2\\ =\dfrac{9}{25}\)
\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3=\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3=\dfrac{4}{9}\)
\(27^3:3^2=3^9:3^2=3^7=2187\)
a) Ta có:
\(64^8=\left(2^6\right)^8=2^{6\cdot8}=2^{48}\)
\(16^{12}=\left(2^4\right)^{12}=2^{4\cdot12}=2^{48}\)
\(\Rightarrow64^8=16^{12}\)
b) Ta có:
\(\left(\dfrac{1}{16}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{4\cdot10}=\left(\dfrac{1}{2}\right)^{40}\)
Mà: 50 > 40 => `(1/2)^50<(1/2)^40`
c) Ta có:
\(\left(\dfrac{9}{16}\right)^{100}=\left[\left(\dfrac{3}{4}\right)^2\right]^{100}=\left(\dfrac{3}{4}\right)^{200}\)
Mà: `3/4>2/3=>(3/4)^200>(2/3)^200`
\(^{^{ }}\)a,64^8=16^12
b,(1/16)^10<(1/2)^50
c,(2/3)^200>(9/16)^100
CỦA BẠN ĐÂY NẾU SAI THÌ CHO MÌNH XIN LỖI NHÉ
a; 25 x 53 x \(\dfrac{1}{625}\) x 52
= 52 x 53 x \(\dfrac{1}{5^4}\) x 52
= 55 x \(\dfrac{1}{5^4}\) x 52
= 5 x 52
= 53
a)
\(25\cdot5^3\cdot\dfrac{1}{625}\cdot5^2\\ =\left(5^2\cdot5^3\cdot5^2\right)\cdot\dfrac{1}{625}\\ =5^7\cdot\dfrac{1}{5^4}\\ =5^3\)
b)
\(5^2\cdot3^5\cdot\left(\dfrac{3}{5}\right)^2\\ =5^2\cdot3^5\cdot\dfrac{3^2}{5^2}\\ =3^5\cdot3^2\\ =3^7\)
c)
\(\left(-\dfrac{1}{7}\right)^4\cdot49^2\\ =\dfrac{\left(-1\right)^4}{7^4}\cdot\left(7^2\right)^2\\ =\dfrac{1}{7^4}\cdot7^4\\ =1\)
d)
\(\left(\dfrac{1}{16}\right)^2:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{8}\right)^3\\ =\left[\left(\dfrac{1}{2}\right)^4\right]^2:\left(\dfrac{1}{2}\right)^4\cdot\left[\left(-\dfrac{1}{2}\right)^3\right]^3\\ =\left(\dfrac{1}{2}\right)^8:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot-\left(\dfrac{1}{2}\right)^9\\ =-\left(\dfrac{1}{2}\right)^{13}\)
a; \(x\) : (- \(\dfrac{1}{3}\))3 = \(\dfrac{1}{9}\)
\(x\) : (\(-\dfrac{1}{27}\)) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) x (- \(\dfrac{1}{27}\))
\(x\) = - \(\dfrac{1}{243}\)
Vậy \(x\) = - \(\dfrac{1}{243}\)
b; (\(\dfrac{4}{5}\))5 x \(x\) = (\(\dfrac{4}{5}\))7
\(x\) = (\(\dfrac{4}{5}\))7 : (\(\dfrac{4}{5}\))5
\(x\) = \(\dfrac{4^7}{5^7}\) : \(\dfrac{4^5}{5^5}\)
\(x\) = \(\dfrac{4^7}{5^7}\) x \(\dfrac{5^5}{4^5}\)
\(x\) = \(\dfrac{4^2}{5^2}\)
\(x\) = \(\dfrac{16}{25}\)
Vậy \(x\) = \(\dfrac{16}{25}\)
a; 5 - (- \(\dfrac{5}{11}\) ) + (\(\dfrac{1}{3}\))2 : 3
= 5 + \(\dfrac{5}{11}\) + \(\dfrac{1}{9}\) : 3
= \(\dfrac{55}{11}\) + \(\dfrac{5}{11}\) + \(\dfrac{1}{9}\) x \(\dfrac{1}{3}\)
= \(\dfrac{55}{11}\) + \(\dfrac{5}{11}\) + \(\dfrac{1}{27}\)
= \(\dfrac{60}{11}\) + \(\dfrac{1}{27}\)
= \(\dfrac{1620}{297}\) + \(\dfrac{11}{297}\)
= \(\dfrac{1631}{297}\)
b; 23 + 3 x (\(\dfrac{1}{2}\))0 + (- 2)2 : \(\dfrac{1}{2}\)
= 8 + 3 x 1 + 4 : \(\dfrac{1}{2}\)
= 8 + 3 + 4 x \(\dfrac{2}{1}\)
= 8 + 3 + 8
= 11 + 8
= 19
11 × 68 + 46 × 33
= 11 × 68 + 46 × 3 × 11
= 11 × 68 + 132 × 11
= 11 × (68 + 132)
= 11 × 200
= 11 × 2 × 100
= 22 × 100
= 2200
11 x 68 + 46 x 33
=748 + 1518
= 2266
ko biết đúng ko nữa
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+50\right)=1175\)
\(50x+\left(1+2+...+50\right)=1175\)
\(50x+\dfrac{50\cdot\left(50+1\right)}{2}=1175\)
\(50x+1275=1175\)
\(50x=1175-1275\)
\(50x=-100\)
\(x=-100:50\)
\(x=-2\)
Vậy...
sửa \(\left(x+1\right)+\left(x+2\right)+...+\left(x+50\right)=1175\)
\(\Leftrightarrow50x+1275=1175\Leftrightarrow x=-2\)