\(3x+2\left(5-x\right)=0\)

\(3x+10-2x=0\)

\(x+10=0\)

\(x=-10\)

\(3x+2\left(5-x\right)=0\)

\(\Leftrightarrow3x+10-2x=0\)

\(\Leftrightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

a, \(\left(\frac{x^3+1}{x^2-1}-\frac{x^2-1}{x-1}\right):\left(x+\frac{x}{x-1}\right)\)

\(=\left(\frac{x^3+1}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x\left(x-1\right)}{x-1}+\frac{x}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x\left(x-1\right)+x}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left[x^2-x+1-x^2+1\right]}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2}{x-1}\right)\)

\(=\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{x^2}=\frac{2-x}{x^2}\)

b, Ta có : A = 3 hay  \(\frac{2-x}{x^2}=3\)

\(3x^2=2-x\Leftrightarrow3x^2+x-2=0\)

\(\Leftrightarrow3x^2+3x-2x-2=0\Leftrightarrow\left(x+1\right)\left(3x-2\right)=0\Leftrightarrow x=-1;\frac{2}{3}\)

a,\(A=\left(\frac{x^3+1}{x^2-1}-\frac{x^2-1}{x-1}\right)\div\left(x+\frac{x}{x-1}\right)\)

\(=\left(\frac{x^3+1}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x^2-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\div\left(\frac{x\left(x-1\right)}{x-1}+\frac{x}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\div\left(\frac{x\left(x-1\right)+x}{\left(x-1\right)}\right)\)

\(=\left(\frac{\left(x+1\right)\left(x^2-x+1-x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\right)\div\left(\frac{x^2}{x-1}\right)\)

\(=\left(\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{x^2}{x-1}\)

\(=\frac{\left(x+1\right)\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{x^2}\)

\(=\frac{\left(x+1\right)\left(2-x\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)x^2}=\frac{2-x}{x^2}\)

giúp mình nhanh với

giúp mình nhanh với

\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right)\times\frac{x^2+8x+16}{32}\)

ĐKXĐ : \(x\ne\pm4\)

\(=\left(\frac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\right)\times\frac{\left(x+4\right)^2}{32}\)

\(=\left(\frac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}\right)\times\frac{\left(x+4\right)^2}{32}\)

\(=\frac{32}{\left(x-4\right)\left(x+4\right)}\times\frac{\left(x+4\right)^2}{32}\)

\(=\frac{x+4}{x-4}\)

\(\left(\frac{4}{x-4}-\frac{4}{x+4}\right).\frac{x^2+8x+16}{32}\)

\(=\left(\frac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\right).\frac{\left(x+4\right)^2}{32}\)

\(=\frac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}.\frac{\left(x+4\right)^2}{32}=\frac{32}{\left(x-4\right)\left(x+4\right)}.\frac{\left(x+4\right)^2}{32}=\frac{x+4}{x-4}\)