\(|x+\dfrac{1}{3}|=\dfrac{3}{4}\)
\(x+\dfrac{1}{3}=\left\{{}\begin{matrix}\dfrac{3}{4}\\-\dfrac{3}{4}\end{matrix}\right.\)
\(x=\left\{{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{3}\\-\dfrac{3}{4}-\dfrac{1}{3}\end{matrix}\right.\)
\(x=\left\{{}\begin{matrix}\dfrac{5}{12}\\\dfrac{-13}{12}\end{matrix}\right.\)
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(5^2.x^2+1^2=\dfrac{36}{49}\)
\(25.x^2+1=\dfrac{36}{49}\)
\(25.x^2=\dfrac{36}{49}+1\)
\(25.x^2=\dfrac{85}{49}\)
\(x^2=\dfrac{85}{49}:25\)
\(x^2=\dfrac{17}{245}\)
\(x\approx\dfrac{4}{16}\approx\dfrac{1}{4}\)
e: \(\left|x+\dfrac{1}{3}\right|=\dfrac{3}{4}\)
=>\(\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{3}{4}\\x+\dfrac{1}{3}=-\dfrac{3}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{4}-\dfrac{1}{3}=\dfrac{5}{12}\\x=-\dfrac{3}{4}-\dfrac{1}{3}=-\dfrac{13}{12}\end{matrix}\right.\)
h: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
=>\(\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}5x=\dfrac{6}{7}-1=-\dfrac{1}{7}\\5x=-\dfrac{6}{7}-1=-\dfrac{13}{7}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{7}:5=-\dfrac{1}{35}\\x=-\dfrac{13}{7}:5=-\dfrac{13}{35}\end{matrix}\right.\)
f: \(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
=>\(\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
=>\(\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=0\\2x=-\dfrac{3}{5}-\dfrac{3}{5}=-\dfrac{6}{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\end{matrix}\right.\)
i: \(2-\sqrt{x}=-7\)
=>\(\sqrt{x}=2+7=9\)
=>\(x=9^2=81\left(nhận\right)\)
