\(a,x^2-10x-39=0\)
\(\Leftrightarrow x^2-10x-39+64=64\)
\(\Leftrightarrow x^2-10x+25=64\)
\(\Leftrightarrow\left(x-5\right)^2=64\)
làm nốt
\(x^2-10x-39=0\Leftrightarrow x^2-13x+3x-39=0\Leftrightarrow x\left(x-13\right)+3\left(x-13\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=13\\x=-3\end{cases}}\)
\(b,\frac{x^2}{x^3-9}=\frac{1}{x+3}\)
\(\Leftrightarrow x^2\left(x+3\right)=x^3-9\)
\(\Leftrightarrow x^3+3x^2=x^3-9\)
\(\Leftrightarrow3x^2=-9\left(VL\right)\)
a, x^2-10x-39=0
<=>x^2+3x-13x-39=0
<=>(x^2+3x)-(13x-39)=0
<=>x(x+3)-13(x+3)=0
<=>(x-13)(x+3)=0
<=>\(\hept{\begin{cases}x-13=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=13\\x=-3\end{cases}}}\)
vậy S={-3; 13}
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
\(\Leftrightarrow\frac{2x-2}{4x^2-8x}-\frac{5-x}{4x^2-8x}=\frac{1}{8x-16}-\frac{1}{8x}\)
\(\Leftrightarrow\frac{3x-7}{4x^2-8x}=\frac{8x-8x+16}{8x\left(8x-16\right)}\)
\(\Leftrightarrow\frac{16\left(3x-7\right)}{16\left(4x^2-8x\right)}=\frac{16}{64x^2-128x}\)
\(\Leftrightarrow48x-112=16\)
\(\Leftrightarrow48x=128\)
\(\Leftrightarrow x=\frac{128}{48}\)(sai đừng ném đá mak ib vs mik nha)
