Bài 2:
a, \(x^{16}-1=\left(x^8\right)^2-1^2\)
\(=\left(x^8-1\right)\left(x^8+1\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
b, \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
Chúc bạn học tốt!!!
Bài 1:
a, \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Rightarrow2x-5=0\Rightarrow x=\dfrac{5}{2}\)
b, \(2x^3+3x^2-2x-3=0\)
\(\Rightarrow2x^3-2x^2+5x^2-5x+3x-3=0\)
\(\Rightarrow2x^2\left(x-1\right)+5x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x^2+5x+3\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x^2+2x+3x+3\right)=0\)
\(\Rightarrow\left(x-1\right)\left[2x\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c, \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+4x\right)=0\)
\(\Rightarrow x\left(x+3\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-4\end{matrix}\right.\)
Chúc bạn học tốt!!!
bài 1) a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow4x^2-25-\left(4x^2+14x-10x-35\right)=0\)
\(\Leftrightarrow4x^2-25-4x^2-14x+10x+35=0\)
\(\Leftrightarrow-4x+10=0\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{10}{4}\) vậy \(x=\dfrac{10}{4}\)
b) \(2x^3+3x^2-2x-3=0\Leftrightarrow\left(2x^3-2x\right)+\left(3x^2-3\right)=0\)
\(\Leftrightarrow2x\left(x^2-1\right)+3\left(x^2-1\right)=0\Leftrightarrow\left(2x+3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}2x+3=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=-3\\x=-1\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{2}\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-3}{2};x=-1;x=1\)
c) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x^2-3x+9+x-9\right)\left(x+3\right)=0\Leftrightarrow\left(x^2-2x\right)\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
vậy \(x=0;x=2;x=-3\)
bài 2) a) \(x^{16}-1=\left(x^8\right)^2-1=\left(x^8\right)^2-1^2=\left(x^8-1\right)\left(x^8+1\right)\)
\(=\left(\left(x^4\right)^2-1^2\right)\left(x^8+1\right)=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(\left(x^2\right)^2-1^2\right)\left(x^4+1\right)\left(x^8+1\right)=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\)
b) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
