a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Rightarrow5x+1=\dfrac{6}{7}\) hoặc \(5x+1=-\dfrac{6}{7}\)
*) \(5x+1=\dfrac{6}{7}\)
\(5x=\dfrac{6}{7}-1\)
\(5x=-\dfrac{1}{7}\)
\(x=-\dfrac{1}{7}:5\)
\(x=-\dfrac{1}{35}\)
*) \(5x+1=-\dfrac{6}{7}\)
\(5x=-\dfrac{6}{7}-1\)
\(5x=-\dfrac{13}{7}\)
\(x=-\dfrac{13}{7}:5\)
\(x=-\dfrac{13}{35}\)
Vậy \(x=-\dfrac{13}{35}\); \(x=-\dfrac{1}{35}\)
b) \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left[\left(\dfrac{2}{3}\right)^2\right]^3\)
\(\left(x-\dfrac{2}{3}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{3}=\dfrac{4}{9}\)
\(x=\dfrac{4}{9}+\dfrac{2}{3}\)
\(x=\dfrac{14}{9}\)
Vậy \(x=\dfrac{14}{9}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(8x-1=5\)
\(8x=5+1\)
\(8x=6\)
\(x=\dfrac{6}{8}\)
\(x=\dfrac{3}{4}\)
Vậy \(x=\dfrac{3}{4}\)
d) \(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)
Do \(\left(x-3,5\right)^2\ge0\)
\(\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4=0\)
\(\Rightarrow\left(x-3,5\right)^2=0\) và \(\left(y-\dfrac{1}{10}\right)^4=0\)
*) \(\left(x-3,5\right)^2=0\)
\(x-3,5=0\)
\(x=0+3,5\)
\(x=3,5\)
*) \(\left(y-\dfrac{1}{10}\right)^4=0\)
\(y-\dfrac{1}{10}=0\)
\(y=0+\dfrac{1}{10}\)
\(y=\dfrac{1}{10}\)
Vậy \(x=3,5\); \(y=\dfrac{1}{10}\)
a) \(\left(5x+1\right)^2=\dfrac{36}{49}< =>\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2\)
\(< =>\left(5x+1\right)^2-\left(\dfrac{6}{7}\right)^2=0< =>\left(5x+1-\dfrac{6}{7}\right)\left(5x+1+\dfrac{6}{7}\right)=0\)
\(< =>\left(5x+\dfrac{1}{7}\right)\left(5x+\dfrac{13}{7}\right)=0\)
\(< =>\left\{{}\begin{matrix}5x+\dfrac{1}{7}=0\\5x+\dfrac{13}{7}=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
b) \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6< =>\left(x-\dfrac{2}{9}\right)^3=\left[\left(\dfrac{2}{3}\right)^2\right]^3\)
\(< =>x-\dfrac{2}{9}=\dfrac{4}{9}< =>x=\dfrac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}< =>8x-1=5< =>x=\dfrac{3}{4}\)
d) \(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0< =>\left\{{}\begin{matrix}\left(x-3,5\right)^2=0\\\left(y-\dfrac{1}{10}\right)^4=0\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}\end{matrix}\right.\)
