Bài 6: Lũy thừa của một số hữu tỉ (tiếp theo...)

Có \(\dfrac{a}{b}=\dfrac{c}{d}< =>\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

<=> \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\left(\dfrac{a+b}{c+d}\right)^2\)

<=> \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2\)(1)

Có \(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)

Áp dụng DTSBN ta có: 

\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\)(2)

Từ (1) (2) => đpcm.

Ta có: \(3\cdot A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

Do đó: 

\(3\cdot A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)

hay \(2\cdot A=1-\dfrac{1}{3^{100}}\)

\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right):2\)

\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right)\cdot\dfrac{1}{2}\)

\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{100}}< \dfrac{1}{2}\)

hay A<B

⇔A=12−12⋅3100<12⇔A=12−12⋅3100<12

hay A<B

\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=\dfrac{1}{100}\)

\(x.5=x^2\)

\(\Rightarrow x^2-5x=0\)

\(\Rightarrow x.\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Vậy \(x\in\left\{0;5\right\}\)

\(\left(3x-1\right)^{2017}=\left(3x-1\right)^{2018}\)

\(\Rightarrow\left(3x-1\right)^{2018}-\left(3x-1\right)^{2017}=0\)

\(\Rightarrow\left(3x-1\right)^{2017}.\left[\left(3x-1\right)-1\right]=0\)

\(\Rightarrow\left(3x-1\right)^{2017}.\left(3x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{2017}=0\\\left(3x-2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\)

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

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

Vậy \(x\in\left\{\dfrac{1}{3};\dfrac{2}{3}\right\}\)

\(\left(x-1\right)^{x+2}=\left(x-1\right)^x\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^x\)

\(\Rightarrow\left(x-1\right)^x.\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^x=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{0;1;2\right\}\)

5⁵.4.10^-5=5⁵.2².5^-5.2^-5=1.2^-3=\(\dfrac{1}{8}\)

a, (5^-5)^-1 . \(\left(\dfrac{1}{2}\right)^{-2}\) . \(\dfrac{1}{10^5}\)

= 5⁵ . \(\dfrac{1^{-2}}{2^{-2}}\) . \(\dfrac{1}{10^5}\)

= (5⁵ . \(\dfrac{1}{10^5}\)) . \(\dfrac{1}{4}\)

= \(\dfrac{5^5}{10^5}\) . \(\dfrac{1}{4}\) = \(\left(\dfrac{1}{2}\right)^5\). \(\dfrac{1}{4}\)

= \(\dfrac{1}{32}.\text{​​}\dfrac{1}{4}\)= \(\dfrac{1}{128}\)

b: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{4}{5}\)

c: \(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^3\cdot5^2}=10\cdot7=70\)

\(\dfrac{3x+2}{4}=\dfrac{2y+2}{5}\)\(=\dfrac{3x+2y+4}{4,5x}\)

= \(\dfrac{3x+2+2y+2-\left(3x+2y+4\right)}{4+5-4,5x}\)

= \(\dfrac{3x+2+2y+2-3x-2y-4}{4+5-4,5x}\)

= \(\dfrac{0}{9-4,5x}\) = 0

Giải tiếp cho bạn Nguyễn Linh nhé :

\(\Rightarrow\left\{{}\begin{matrix}3x+2=0\cdot4=0\\2y+2=0\cdot5=0\end{matrix}\right.\)

\(\Rightarrow3x+2=2y+2\)

\(\Rightarrow3x=2y\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\) . Từ đây bạn áp dụng điều kiện thứ 2 của đề bài để tính x và y nhé

a)-5^x.5^x+4.5^x+3+(-4)-3=2.5¹¹

5^x+6.5¹¹-2=4+3

5^x+17=7-2

5^x+17=5¹

=>x+17=1

=>x=-16

mk bt câu a thôi

b: \(\Leftrightarrow5^x\cdot75+5^x\cdot\dfrac{4}{125}=19\cdot5^{10}\)

\(\Leftrightarrow5^x\simeq24729\)

hay \(x\in\varnothing\)

c: \(\Leftrightarrow2^x\cdot\left(5+3\cdot4\right)=6^{11}\left(11+2\cdot36\right)\)

=>2^x=1771303273

hay \(x\in\varnothing\)

thi chac chan 334<520 rui

334<520

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