Bài 7: Lũy thừa với số mũ tự nhiên. Nhân hai lũy thừa cùng cơ số. Luyện tập

nhìn thoy đã thấy nản r`....

a/ \(3^{500}=\left(3^5\right)^{100}=243^{100};5^{300}=\left(5^3\right)^{100}=125^{100}\)

Ta thấy \(243^{100}>125^{100}\Rightarrow3^{500}>5^{300}\)

b/ \(125^5=\left(5^3\right)^5=5^{15};25^7=\left(5^2\right)^7=5^{14}\)

ta thấy \(5^{15}>5^{14}\Rightarrow125^5>25^7\)

c/ \(9^{20}=\left(3^2\right)^{20}=3^{40};27^{13}=\left(3^3\right)^{13}=3^{39}\)

Ta thấy \(3^{40}>3^{39}\Rightarrow9^{20}>27^{13}\)

...còn lại tự lm nốt nhá....

Ta có :

\(31^{11}< 32^{11}=16^{11}.2^{11}\)

\(16^{11}.2^{12}=16^{14}< 17^{14}\)

Lại có :

\(16^{11}.2^{11}< 16^{11}.2^{12}\)

\(\Rightarrow31^{11}< 16^{11}.2^{11}< 16^{11}.2^{12}< 17^{14}\)

\(\Rightarrow31^{11}< 17^{14}\)

\(31^{11}< 17^{14}\)

Ta có:

\(31^{11}< 32^{11}\Leftrightarrow31^{11}< \left(2^5\right)^{11}\Leftrightarrow31^{11}< 2^{55}\left(1\right)\)

\(17^{14}>16^{14}\Leftrightarrow17^{14}>\left(2^4\right)^{14}\Leftrightarrow17^{14}>2^{56}\left(2\right)\)

Từ (1) và (2) ta có:

\(31^{11}< 17^{14}\)

\(2^3.\dfrac{1}{6}=8.\dfrac{1}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

\(2^3\cdot\dfrac{1}{6}\)

\(8\cdot\dfrac{1}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

\(2^3\) .\(\dfrac{1}{6}\)=8 .\(\dfrac{1}{6}\)=\(\dfrac{8}{6}\)=1\(\dfrac{1}{3}\)

đề sai r :v

có mũ ko v:v

\(x^{2016}=x^{2017}\)

\(\Rightarrow x^{2016}-x^{2017}=0\)

\(\Rightarrow x^{2016}.\left(1-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^{2016}=0\\1-x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy........

Chúc bạn học tốt!!!

\(x^{2016}=x^{2017}\)

\(\Leftrightarrow0=x^{2017}-x^{2016}\)

\(\Leftrightarrow0=x^{2016}\left(x-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x^{2016}=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)\(\left(TM\right)\)

Vậy \(x\in\left\{0;1\right\}\) là giá trị cần tìm

\(x^{2016}=x^{2017}\)

\(\Rightarrow x^{2017}-x^{2016}=0\)

\(x^{2016}\left(x-1\right)=0\)

\(x^{2016}=0\Rightarrow x=0\)

\(x-1=0\Rightarrow x=1\)

a, A= { x\(\in\)N | 0\(\le\)x\(\le\)2017}

chỉ giúp vậy thôi

3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)

\(\Leftrightarrow2x-6-3+6x=4+4-4x\)

\(\Leftrightarrow8x-9=8-4x\)

\(\Leftrightarrow8x+4x=8+9\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\dfrac{17}{12}\)

Vậy \(x=\dfrac{17}{12}\)

4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)

\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)

\(\Leftrightarrow6x-12-4-4x=12-9x-12\)

\(\Leftrightarrow6x-4-4x=12-9x\)

\(\Leftrightarrow2x-4=12-9x\)

\(\Leftrightarrow2x+9x=12+4\)

\(\Leftrightarrow11x=16\)

\(\Leftrightarrow x=\dfrac{16}{11}\)

Vậy \(x=\dfrac{16}{11}\)