a) \({( - 2)^4} \cdot {( - 2)^5} = {\left( { - 2} \right)^{4 + 5}} = {\left( { - 2} \right)^9}\)

 \({( - 2)^{12}}:{( - 2)^3} = {\left( { - 2} \right)^{12 - 3}} = {\left( { - 2} \right)^9}\)

Vậy \({( - 2)^4} \cdot {( - 2)^5}\) = \({( - 2)^{12}}:{( - 2)^3}\);

b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)

\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)

Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)

c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 - 2}} = {\left( {0,3} \right)^6}\)

\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)

Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).

d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3} = {\left( { - \frac{3}{2}} \right)^{5 - 3}} = {\left( { - \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)

Vậy \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).

(-2) ^4 . (-2) 65 và ( -2) ^ 12 : ( -2) ^3

=( -2) ^ 4+5 =(-2)^9 và (-2) ^12-3 = ( -2) ^9 

vậy ( -2) ^9 = (-2) ^9 

Nên (-2) ^4 .( -2) ^5 = ( -2) ^ 12 : ( -2) ^3

Câu 1 :

a) \(4.\left(\frac{1}{32}\right)^{-2}:\left(2^3.\frac{1}{16}\right)\)

\(=2^2.32^2:\left(\frac{1}{8}.16\right)=\left(2.32\right)^2:2=64^2:2\)

\(=2048=2^{11}\)

b) \(5^2.3^5.\left(\frac{3}{5}\right)^2\)

\(=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)

VIẾT CÁC BIỂU THỨC DƯỚI DẠNG LUỸ THỪA CỦA 1 SỐ HỮU TỈ

\(a,4\cdot\left(\frac{1}{32}\right)^{-2}:\left(2^3\cdot\frac{1}{16}\right)\\ =4\cdot1024:\left(8\cdot\frac{1}{16}\right)\\ =4\cdot1024:\frac{1}{2}\\ =2\cdot1024\\ =2\cdot2^{10}\\ =2^{11}\)

\(b,5^2\cdot3^5\cdot\left(\frac{3}{5}\right)^2\\ =5^2\cdot\left(\frac{3}{5}\right)^2\cdot3^5\\ =3^2\cdot3^5\\ =3^7\)

2 SO SÁNH

\(a,10^{20}\text{ và }9^{10}\)

Có: \(9^{10}=\left(3^2\right)^{10}=3^{20}\)

\(\Rightarrow10^{20}>3^{20}\\ \text{hay}\text{ }10^{20}>9^{10}\)

\(b,\left(-5\right)^3\text{ và }\left(-3\right)^{50}\)

Có: \(\left(-3\right)^{50}=3^{50}\)

\(\Rightarrow\left(-5\right)^3< 3^{50}\\ \text{hay }\left(-5\right)^3< \left(-3\right)^{50}\)

\(c,64^3\text{ và }16^{12}\)

Có: \(64^3=\left(4^3\right)^3=4^9;16^{12}=\left(4^2\right)^{12}=4^{24}\)

\(\Rightarrow4^9< 4^{24}\\ hay\text{ }64^3< 16^{12}\)

\(d,\left(\frac{1}{16}\right)^{10}\text{ và }\left(\frac{1}{2}\right)^{50}\)

Có: \(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5\cdot10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\\ \text{hay }\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

a) Vì \(\dfrac{1}{24}< \dfrac{1}{83}\) 

⇒ \(\dfrac{1}{24^9}>\dfrac{1}{83^{13}}\)

a) \(\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{27}\right)^9=\dfrac{1}{3^{27}}\)

\(\left(\dfrac{1}{83}\right)^{13}< \left(\dfrac{1}{81}\right)^{13}=\dfrac{1}{3^{52}}\)

Mà \(\dfrac{1}{3^{27}}>\dfrac{1}{3^{52}}\)

\(\Rightarrow\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{83}\right)^{13}\)

b) \(3^{300}=\left(3^3\right)^{100}=27^{100}\)

\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\)

Mà \(25^{100}< 27^{100}\)

\(\Rightarrow5^{199}< 3^{300}\)

\(\Rightarrow\dfrac{1}{5^{199}}>\dfrac{1}{3^{300}}\)

\(a,\left(\dfrac{1}{24}\right)^9=\dfrac{1}{24^9};\left(\dfrac{1}{83}\right)^{13}=\dfrac{1}{83^{13}};24^9< 83^{13}\left(24< 83;9< 13\right)\\ \Rightarrow\dfrac{1}{24^9}< \dfrac{1}{83^{13}}\Rightarrow\left(\dfrac{1}{24}\right)^9< \left(\dfrac{1}{83}\right)^{13}\\ b,3^{300}=27^{100}>25^{100}=5^{200}>5^{199}\\ \Rightarrow\dfrac{1}{3^{300}}< \dfrac{1}{5^{199}}\)

Ta có:

(-1/5)³⁰⁰ = (-1)³⁰⁰/5³⁰⁰ = 1/(5³)¹⁰⁰ = 1/125¹⁰⁰

(-1/3)⁵⁰⁰ = (-1)⁵⁰⁰/3⁵⁰⁰ = 1/(3⁵)¹⁰⁰ = 1/243¹⁰⁰

Vì 125¹⁰⁰ < 243¹⁰⁰

=> 1/125¹⁰⁰ > 1/243¹⁰⁰

=> (-1/5)³⁰⁰ > (-1/3)⁵⁰⁰

Ta có : \(\left(-\frac{1}{5}\right)^{300}=\left(-\frac{1}{5}\right)^{3.100}=\left(-\frac{1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

            \(\left(-\frac{1}{3}\right)^{500}=\left(-\frac{1}{3}\right)^{5.100}=\left(-\frac{1}{243}\right)^{100}=\left(\frac{1}{243}\right)^{100}\)

Mà \(125< 243\Rightarrow\frac{1}{125}>\frac{1}{243}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{243}\right)^{100}\)

\(=>\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)

Ta có:

(-1/5)³⁰⁰ = (-1)³⁰⁰/5³⁰⁰ = 1/(5³)¹⁰⁰ = 1/125¹⁰⁰

(-1/3)⁵⁰⁰ = (-1)⁵⁰⁰/3⁵⁰⁰ = 1/(3⁵)¹⁰⁰ = 1/243¹⁰⁰

Vì 125¹⁰⁰ < 243¹⁰⁰

=> 1/125¹⁰⁰ > 1/243¹⁰⁰

=> (-1/5)³⁰⁰ > (-1/3)⁵⁰⁰

Ta có :

M = \(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

M = \(\frac{1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{91}+1\right)+...+\left(\frac{98}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

M = \(\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

M = \(\frac{100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

M = \(100\)

N = \(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)

N = \(\frac{\left(1-\frac{1}{9}\right)+\left(1-\frac{2}{10}\right)+\left(1-\frac{3}{11}\right)+...+\left(1-\frac{92}{100}\right)}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)

N = \(\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)

N = \(\frac{8.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{5}.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}\)

N = \(40\)

\(\Rightarrow\)M : N = \(\frac{100}{40}\%=250\%\)

thiếu đề r bn

\(M=\frac{1+(\frac{1}{99}+1)+(\frac{2}{98}+1)+(\frac{3}{97}+1)+...+(\frac{98}{2}+1)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(M=\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

\(M=\frac{100\cdot(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2})}{(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})}=100\)

\(N=\frac{(1-\frac{1}{9})+(1-\frac{2}{10})+(1-\frac{3}{11})+...+(1-\frac{92}{100})}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)

\(N=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}=\frac{8(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100})}{\frac{1}{5}(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100})}=40\)

\(M:N=\frac{100}{40}=250\%\)

a: \(6\sqrt{3}=\sqrt{108}>\sqrt{54}=3\sqrt{6}\)

\(\Rightarrow5^{6\sqrt{3}}>5^{3\sqrt{6}}\)

b: \(\sqrt{2}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}+\dfrac{2}{3}}=2^{\dfrac{7}{6}}\)

\(\left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}=2^{\left(-1\right)\cdot\left(-\dfrac{4}{3}\right)}=2^{\dfrac{4}{3}}\)

mà \(\dfrac{7}{6}< \dfrac{8}{6}=\dfrac{4}{3}\).

nên \(\sqrt{2}\cdot2^{\dfrac{2}{3}}< \left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}\).

a) Ta có :\(\left(\frac{-1}{5}\right)^{300}=\frac{-1^{300}}{5^{300}}=\frac{1}{125^{100}}\)

\(\left(\frac{-1}{3}\right)^{500}=\frac{-1^{500}}{3^{500}}=\frac{1}{243^{100}}\)

Mà \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\)

\(\Rightarrow\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b)Ta có :\(2^{90}=\left(2^{15}\right)^6=32768^6\)

\(5^{36}=\left(5^6\right)^6=15625^6\)

Vì \(32768^6>15625^6\Rightarrow2^{90}>5^{36}\)

a.Ta có: \(\left(\frac{-1}{5}\right)^{300}=\left(\frac{-1}{5}^3\right)^{100}=\left(\frac{-1}{125}\right)^{100}=\left(\frac{1}{125}\right)^{100}\)

\(\left(\frac{-1}{3}\right)^{500}=\left(\frac{-1}{3}^5\right)^{100}=\left(\frac{-1}{243}\right)^{100}=\left(\frac{1}{234}\right)^{100}\)

Mà: \(\frac{1}{125}>\frac{1}{234}\Rightarrow\left(\frac{1}{125}\right)^{100}>\left(\frac{1}{234}\right)^{100}\)

Vậy \(\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)

b.Ta có: \(2^{90}=\left(2^{10}\right)^9=1024^9\)

\(5^{36}=\left(5^4\right)^9=625^9\)

Mặt khác: \(1024>625\Rightarrow1024^9>625^9\)

 Vậy \(2^{90}>5^{36}\)

#)Giải :

\(\left(92-\frac{1}{9}-\frac{2}{10}-\frac{3}{10}-...-\frac{92}{100}\right):\left(\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}\right)\)

\(=\left(1-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+...+1-\frac{92}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

\(=\left(\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

\(=8\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\div\frac{1}{5}\times\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)\)

\(=8\div\frac{1}{5}\)

\(=40\)

                         #~Will~be~Pens~#

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Chắc đấy! Tick nhé!