\(\dfrac{2}{3}.x=\dfrac{-1}{8}\\<=>x=\dfrac{-1}{8}:\dfrac{2}{3}\\<=>x= \dfrac{-1}{8}.\dfrac{3}{2}\\<=>x=\dfrac{-3}{16}\)

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\(-x+\dfrac{1}{7}=1,5\\<=>-x=1,5-\dfrac{1}{7}\\<=>-x= \dfrac{19}{14}\\<=>x=\dfrac{-19}{14}\)

\(x+\dfrac{1}{2}= \dfrac{3}{4}\\<=>x=\dfrac{3}{4}-\dfrac{1}{2}\\<=>x=\dfrac{3}{4}-\dfrac{2}{4}\\<=>x=\dfrac{1}{4}\)

\(a)\dfrac{6x^2y}{4xy^3}=\dfrac{3x}{2y^2}\\b)\dfrac{4x^2y(x-y)}{12xy^3(x-y)}=\dfrac{x}{3y}\\\)

\(c)\dfrac{5x^2-5x}{x-1}=\dfrac{5x(x-1)}{x-1}=5x\\\)

\(d)\dfrac{3x^2-9x}{xy(x-2)}=\dfrac{3x(x-3)}{xy(x-2)}\\=\dfrac{3(x-3)}{y(x-2)}\)

\(a)7x(x-5)-x+5=0\\<=>7x(x-5)-(x-5)=0\\<=>(x-5)(7x-1)=0\\<=> \left[\begin{matrix} x-5=0\\ 7x-1=0\end{matrix}\right.\\<=> \left[\begin{matrix} x=5\\ x=1/7\end{matrix}\right.\\b)(x-3)^2-4(x+2)^2=0\\<=>(x-3)^2-[2x(x+2)]^2=0\\<=>(x-3)^2-(2x^2+4x)^2=0\\<=>(x-3-2x^2-4x)(x-3+2x^2+4x)=0\\<=>(-2x^2-5x-3)(2x^2+5x-3)=0\\<=>(-2x^2-2x-3x-3)(2x^2-x+6x-3)=0\\\)
\(<=>[-2x(x+1)-3(x+1)][x(2x-1)+3(2x-1)]\\<=>(-2x-3)(x+1)(2x-1)(x+3)=0\\TH1: -2x-3=0\\<=>x=3/2\\TH2:x+1=0\\<=>x=-1\\TH3:2x-1=0\\<=>x=1/2\\TH4:x+3=0\\<=>x=-3\)

\(a)x(x+2)=3x+6\\<=>x(x+2)-(3x+6)=0\\<=>x(x+2)-3(x+2)=0\\<=> (x+2)(x-3)=0\\<=>\left[\begin{matrix} x+2=0\\ x-3=0\end{matrix}\right.\\<=> \left[\begin{matrix} x=-2\\ x=3\end{matrix}\right.\\b)4x^2-9+(2x-3)^2=0\\<=>(2x-3)(2x+3)+(2x-3)^2=0\\<=>(2x-3)(2x+3+2x+3)=0\\<=> (2x-3)(4x+6)=0\\<=>\left[\begin{matrix} 2x-3=0\\ 4x+6=0\end{matrix}\right.\\<=>x=3/2 hoặc -3/2\)