Question

[dang van haong]

cac ban giai ho minh nha

Answer 1

[Nguyễn Huy Tú]

a, \(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x+4}{x^2-2x-3}\)ĐK : \(x\ne-1;3\)

\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x+4}{\left(x+1\right)\left(x-3\right)}\)

\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x+8\)

\(\Leftrightarrow x^2+x+x^2-3x=4x+8\Leftrightarrow2x^2-6x-8=0\)

\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x+1\right)\left(x-4\right)=0\Leftrightarrow x=-1\left(ktm\right);x=4\)

b, \(\frac{2x+19}{5x^2-5}-\frac{17}{x^2-1}=\frac{3}{1-x}\)ĐK : \(x\pm1\)

\(\Leftrightarrow\frac{2x+19}{5\left(x-1\right)\left(x+1\right)}-\frac{17}{\left(x-1\right)\left(x+1\right)}=-\frac{15}{5\left(x-1\right)}\)

\(\Rightarrow2x+19-85=-15\left(x+1\right)\)

\(\Leftrightarrow2x-64=-15x-15\Leftrightarrow17x=49\Leftrightarrow x=\frac{49}{17}\)

Answer 2

[Nguyễn Huy Tú]

c, \(x^2-5x+5=0\Leftrightarrow x^2-2.\frac{5}{2}x+\frac{25}{4}-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\)

TH1 : \(x-\frac{5}{2}=\frac{\sqrt{5}}{2}\Leftrightarrow x=\frac{\sqrt{5}+5}{2}\)

TH2 : \(x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\Leftrightarrow x=\frac{5-\sqrt{5}}{2}\)

c, \(2x^3+6x^2=x^2+3x\Leftrightarrow2x^3+5x^2-3x=0\)

\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\Leftrightarrow x=0;x=-3;x=\frac{1}{2}\)