Bài 1:tìm x,biết
a)x3-6x2-x+30=0
\(\Leftrightarrow\text{x³ + 2x² - 8x² - 16x + 15x + 30 = 0
}\)
\(\Leftrightarrow\text{x² ( x + 2 ) - 8x ( x + 2 ) + 15 ( x + 2 ) = 0 }\)
\(\Leftrightarrow\text{( x + 2 )( x² - 8x + 15 ) = 0 }\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+16-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[\left(x-4\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-4-1\right)\left(x-4+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\\x=3\end{matrix}\right.\)
b)x2+16=10x
\(\Leftrightarrow x^2-10x+16=0\)
\(\Leftrightarrow x^2-2x-8x+16=0\)
\(\Leftrightarrow x\left(x-2\right)-8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)
Bài 2:
a)(x-2)2-(x-3)(x-1)\(=x^2-4x+4-\left(x^2-x-3x+3\right)\)
=\(x^2-4x+4-x^2+4x-3\)
\(=1\).Vậy biểu thức a không phụ thuộc vào biến x
b)(x-1)3-(x+1)3+6(x+1)(x-1)
\(=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)
\(=-8\).Vậy biểu thức b không phụ thuộc vào biến x
