1,
a, \(\left(x-\dfrac{1}{7}\right)^4=\left(x-\dfrac{1}{7}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{1}{7}\right)^4-\left(x-\dfrac{1}{7}\right)^2=0\)
\(\Leftrightarrow\left[\left(x-\dfrac{1}{7}\right)^2+x-\dfrac{1}{7}\right]\left[\left(x-\dfrac{1}{7}\right)^2-x+\dfrac{1}{7}\right]=0\)
\(\Leftrightarrow\left[x^2+\dfrac{1}{49}-\dfrac{2}{7}x+x-\dfrac{1}{7}\right]\left[x^2+\dfrac{1}{49}-\dfrac{2}{7}x-x+\dfrac{1}{7}\right]=0\)
\(\Leftrightarrow\left(x^2+\dfrac{5}{7}x-\dfrac{6}{49}\right)\left(x^2-\dfrac{9}{7}x+\dfrac{8}{49}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+\dfrac{5}{7}x-\dfrac{6}{49}=0\\x^2-\dfrac{9}{7}x+\dfrac{8}{49}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=\dfrac{8}{7}\end{matrix}\right.\)
Vậy...
b, \(\left|x+6,4\right|+\left|x+2,5\right|+\left|x+8,1\right|=4x\)
\(\Leftrightarrow x+6,4+x+2,5+x+8,1=4x\) với mọi x
\(\Leftrightarrow x+x+x-4x=-8,1-2,5-6,4\)
\(\Leftrightarrow-x=-17\)
\(\Leftrightarrow x=17\)
Vậy...
