Bài 5: Lũy thừa của một số hữu tỉ

_{\(A=1+2+2^2+2^3+...+2^{10}\)}

_{\(2A=2+2^2+2^3+2^4+...+2^{11}\)}

^{\(2A-A=\left(2+2^2+2^3+2^4+...+2^{11}\right)-\left(1+2+2^2+2^3+...+2^{10}\right)\)}

^{\(A=2^{11}-1< 2^{11}\)}

^{\(B=2.2^2+3.2^3+4.2^4+...+10.2^{10}\)\(2B=2.2^3+3.2^4+4.2^5+...+10.2^{11}\)\(2B-B=\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+\left(9.2^{10}-10.2^{10}\right)+10.2^{11}-2.2^2\)\(B=2^3\left(2-3\right)+2^4\left(3-4\right)+...+2^{10}\left(9-10\right)+10.2^{11}-2.2^2\)\(B=-2^3-2^4-....-2^{10}+10.2^{11}-2^3\)}

_{\(B=-\left(2^3+2^4+...+2^{10}\right)+10.2^{11}-2^3\)}

^{\(B=-\left(2^{11}-2^3\right)+10.2^{11}-2^3\)}

_{\(B=-2^{11}+2^3+10.2^{11}-2^3\)}

_{\(B=9.2^{11}\)}

Ta cần so sánh: _\(9.2^{11}\) và _\(2^{14}\)

Hay _\(9\) và _\(2^3\)

_{\(9>8=2^3\Leftrightarrow B>2^{14}\)}

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

\(=\left(-\dfrac{5}{21}\right):\dfrac{4}{5}+\left(\dfrac{5}{21}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\)

\(=0:\dfrac{4}{5}\)

\(=0\)

b) \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{5}{9}:\left(-\dfrac{3}{22}\right)+\dfrac{5}{9}:\left(-\dfrac{3}{5}\right)\)

\(=\dfrac{5}{9}:\left[\left(-\dfrac{3}{22}\right)+\left(-\dfrac{3}{5}\right)\right]\)

\(=\dfrac{5}{9}:\left(-\dfrac{81}{110}\right)\)

\(=-\dfrac{550}{729}\)

c) \(4^2.4^3:4^{10}\)

\(=\dfrac{4^5}{4^{10}}\)

\(=\dfrac{1}{4^5}\)

\(=\dfrac{1}{256}\)

d) \(\left(0,6\right)^5:\left(0,2\right)^6\)

\(=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^6}\)

\(=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^6}\)

\(=\dfrac{243}{0,2}\)

\(=1215\)

Mai mốt bạn đăng một lần ít thôi nha tại giờ khuya quá nên mình chỉ làm đến đây thôi =))

a.(-5)^8.(-5)^3=(-5)^11

b.(3/8)^7:(3/8)^4=(3/8)^3

\(a)\left(-5\right)^8.\left(-5\right)^3=\left(-5\right)^{8+3}=\left(-5\right)^{11}\)

\(b)\left(\dfrac{3}{8}\right)^7:\left(\dfrac{3}{8}\right)^4=\left(\dfrac{3}{8}\right)^{7-4}=\left(\dfrac{3}{8}\right)^3\)

Chúc bạn học tốt!

(-5)^8 . (-5)^3 = (-5)^8+3=(-5)^11
(3/8)^7:(3/8)^4 = (3/8)^7-4 = (3/8)^3

\(\dfrac{8^{15}.25^7}{125^5.2^{47}}=\dfrac{\left(2^3\right)^{15}.\left(5^2\right)^7}{\left(5^3\right)^5.2^{47}}=\dfrac{2^{45}.5^{14}}{5^{15}.2^{47}}\)

\(\dfrac{2^{45}.5^{14}}{2^{47}.5^{15}}=\dfrac{1.1}{4.5}=\dfrac{1}{20}\)

\(\left|-a^2\right|=\left|-\left(a^2\right)\right|=a^2\)

Vì \(a^2=a^2\)

Vậy \(a^2=\left|-a^2\right|\)

Ta thấy: \(\left|-a\right|^2=a^2\)

Vì \(a^2=a^2\)

\(\Rightarrow a^2=\left|-a\right|^2\)

Vậy ...............

\(32^9=\left(2^5\right)^9=2^{5.9}=2^{45}=2^{13}.2^{32}\)

\(18^{13}=\left(2.3^2\right)^{13}=2^{13}.\left(3^2\right)^{13}=2^{13}.3^{26}\)

\(2^{32}=2^{16.2}=\left(2^{16}\right)^2=65536^2\)

\(3^{26}=3^{13.2}=\left(3^{13}\right)^2=1594323^2\)

Vì \(65536^2< 1594323^2\)

Do đó \(2^{32}< 3^{26}\)

Nên \(2^{13}.2^{32}< 2^{13}.3^{26}\)

Vậy \(32^9< 18^{13}\)

Ta có:

\(32^9=\left(2^5\right)^9=2^{45}\)

\(18^{13}>16^{13}=\left(2^4\right)^{13}=2^{52}\left(^4\right)\)

Ta thấy: \(2^{45}< 2^{52}< 18^{13}\)

\(\Rightarrow32^9< 18^{13}\)

\(32^9\) và \(18^{13}\)

Ta có :

\(32^9=\left(2^5\right)^9=2^{45}\)

\(18^{13}>16^{13}=\left(2^4\right)^{13}=2^{52}\left(^4\right)\)

Ta thấy : \(2^{45}< 2^{52}< 18^{13}\)

\(\Rightarrow32^9< 18^{13}\)

a) \(\left(x-3\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=1^2\\\left(x-3\right)^2=-1^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

b) \(\left(x-\dfrac{1}{7}\right)^2=0\)

\(\Rightarrow x-\dfrac{1}{7}=0\)

\(\Rightarrow x=0+\dfrac{1}{7}\)

\(\Rightarrow x=\dfrac{1}{7}\)

c) \(\left(2x+3\right)^3=-27\)

\(\Rightarrow\left(2x+3\right)^3=\left(-3\right)^3\)

\(\Rightarrow2x+3=-3\)

\(\Rightarrow2x=-6\)

\(\Rightarrow x=-3\)

d) \(-\left(5+35x\right)^2=36\)

\(\Rightarrow\left[{}\begin{matrix}\left(-5-35x\right)^2=6^2\\\left(-5-35x\right)^2=\left(-6\right)^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-5-35x=6\\-5-35x=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}35x=-11\\35x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{35}\\x=\dfrac{1}{35}\end{matrix}\right.\)

a) (x-3)mũ 2 = 1

Vậy x-3 = 1( vì 1 mũ 2 sẽ bằng 1)

=> x = 1+3 = 4

b) (x - 1/7) mũ 2 = 0

Vậy x - 1/7 = 0 ( vì 0 mũ 2 sẽ bằng 0)

=> x = 0 + 1/7 = 1/7

c) (2x + 3 ) mũ 3 = -27

vậy 2x + 3 = -3 ( vì -3 mũ 3 sẽ bằng -27)

=> 2x = -3-3 = -6

=> x = -6/2 = -3

d) ta có :1/2 mũ 2 sẽ = 1/4

=> ( x + 1/2 ) mũ 2 = 1/2 mũ 2

=> x + 1/2 = 1/2 ( 2 bên đều mũ 2 mũ 2 bỏ)

vậy x = 1/2 - 1/2 = 0

Vì \(45=BCNN\left(5,9\right)\) và \(ƯCLN\left(5,9\right)=1\)

Ta có :

\(36^{36}-9^{10}⋮9\left(1\right)\)

Mặt khác :

\(36^{36}=\left(......6\right)\)

\(9^{10}=\left(9^2\right)^5=81^5=\left(.....1\right)\)

\(\Leftrightarrow36^{36}-9^{10}=\left(....6\right)-\left(....1\right)=\left(.....5\right)\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow36^{36}-9^{10}⋮45\left(đpcm\right)\)

0o0^^^Nhi^^^0o0 tưởng bn bảo là bn lp 9 mà sao gửi câu hỏi lp 7 vậy

\(2x^7=5x^5\)

\(\Rightarrow\dfrac{x^7}{x^5}=\dfrac{5}{2}\)

\(\Rightarrow x^{7-5}=\dfrac{5}{2}\)

\(\Rightarrow x^2=\dfrac{5}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{2}}\\x=-\sqrt{\dfrac{5}{2}}\end{matrix}\right.\)

\(5^{3x+3}\le10^{18}:8^6\)
\(\Leftrightarrow5^{3x+3}\le\left(2.5\right)^{18}:\left(2^3\right)^6\)
\(\Leftrightarrow5^{3x+3}\le2^{18}.5^{18}:2^{18}\)
\(\Leftrightarrow5^{3x+3}\le5^{18}\)
\(\Leftrightarrow3x+3\le18\)
\(\Leftrightarrow3x\le15\)
\(\Leftrightarrow x\le5\)
Vậy \(x\le5\)
Chúc bạn học tốt (≧▽≦)