Bài 5: Lũy thừa của một số hữu tỉ

\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}\)

= \(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}\)

= \(1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-1\dfrac{15}{17}\)

= \(1+1-1\dfrac{15}{17}\)

= \(2-1\dfrac{15}{17}\)

= \(\dfrac{2}{17}\)

ta có : \(\left|x+1\right|=\dfrac{3}{7}\Leftrightarrow\left[{}\begin{matrix}x+1=\dfrac{3}{7}\\x+1=\dfrac{-3}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4}{7}\\x=\dfrac{-10}{7}\end{matrix}\right.\)

vậy \(x=\dfrac{-4}{7};x=\dfrac{-10}{7}\)

TH1:x+1=3/7 TH2:x+1=-3/7

x=-4/7 x=-10/7

a) Ta thấy \(\left|x-2\right|\ge0\) ; \(\left|y-1\right|\ge0\)

\(\Rightarrow\left|x-2\right|+\left|y-1\right|\ge0\) mà \(\left|x-2\right|+\left|y-1\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\) Tự tính tiếp

\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}=\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)

Lời giải:

Ta có:

\(71^{15}=(71^3)^5>(68^3)^5=[(17.4)^3]^5=(17^3.64)^5>(17^3.17)^5=(17^4)^5=17^{20}\)

Vậy \(71^{15}> 17^{20}\)

\(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)

\(3^{x-1}.\left(4+2.3^3\right)=3^6.\left(4+2.3^3\right)\)

\(\Leftrightarrow3^{x-1}=3^6\)

\(\Leftrightarrow x-1=6\)

\(\Leftrightarrow x=7\)

Vậy \(x=7\)

\(5^{x+4}-3.5^{x+3}=2.5^{11}\)

\(\Leftrightarrow5^{x+3}.\left(5-3\right)=2.5^{11}\)

\(\Leftrightarrow5^{x+3}.2=2.5^{11}\)

\(\Leftrightarrow5^{x+3}=5^{11}\)

⇔\(x+3=11\)

\(\Leftrightarrow x=8\)

Vậy \(x=8\)

\(11.6^{x-1}=11.6^{11}+2.6^{13}\)

\(\Rightarrow11.6^{x-1}=6^{11}.\left(11+2.36\right)\)

\(\Rightarrow11.6^{x-1}=6^{11}.83\)

\(\left(8x-1\right)^{2n+1}=5^{2n+1}\)

\(Vì\) \(2n+1=2n+1\)

\(\Rightarrow8x-1=5\)

\(\Rightarrow8x=6\)

\(\Rightarrow x=\dfrac{6}{8}=\dfrac{3}{4}\)

\(Vậy\) \(x=\dfrac{3}{4}\)