Phân tích đa thức thành nhân tử:
\(x^2+x-30\)
Tìm x:
a)\(\left(x-2\right)^2-x\left(x-5\right)=13\)
b)\(4x^3-100x=0\)
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\(\frac{4}{x-5}\)- \(\frac{1}{x+5}\)+\(\frac{13x-x^2}{\left(5-x\right)\left(5+x\right)}\)= \(\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)- \(\frac{x-5}{\left(x-5\right)\left(x+5\right)}\)- \(\frac{x\left(x-13\right)}{\left(x-5\right)\left(5+x\right)}\)=\(\frac{4\left(x+5\right)-x+5+x\left(x-13\right)}{\left(x-5\right)\left(x+5\right)}\)= \(\frac{x^2-10x+12}{\left(x-5\right)\left(x+5\right)}\)
\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{\left(5-x\right)\left(x+5\right)}\)ĐKXĐ : \(x\ne\pm5\)
\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x+5\right)\left(x-5\right)}-\frac{13x-x^2}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{4x+20-x+5-13x+x^2}{\left(x-5\right)\left(x+5\right)}=\frac{-10x+25+x^2}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)
\(\Leftrightarrow\)\(\frac{x+1}{x-2}+\frac{8-7x}{x^2-2x}+\frac{2}{x}\left(ĐKXĐ.x\ne2.x\ne0\right)\)
\(\Leftrightarrow\)\(\frac{x\left(x+1\right)}{x\left(x-2\right)}+\frac{8-7x}{x\left(x-2\right)}+\frac{2\left(x-2\right)}{x\left(x-2\right)}\)
\(\Leftrightarrow\frac{x^2+x+8-7x+2x-4}{x\left(x-2\right)}\)
\(\Leftrightarrow\frac{x^2-4x+4}{x\left(x-2\right)}\)
\(\Leftrightarrow\frac{\left(x-2\right)^2}{x\left(x-2\right)}\Leftrightarrow\frac{x-2}{x}\)
Bai 1
\(x^2+x-30=x^2+6x-5x-30=\left(x-5\right)\left(x+6\right)\)
Bai 2
a, \(\left(x-2\right)^2-x\left(x-5\right)=13\)
\(\Leftrightarrow x^2-4x+4-x^2+5x=13\)
\(\Leftrightarrow x+4=13\Leftrightarrow x=9\)
b, \(4x^3-100x=0\Leftrightarrow x\left(4x^2-100\right)=0\)
\(\Leftrightarrow x\left(2x-10\right)\left(2x+10\right)=0\Leftrightarrow x=0;\pm5\)