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Phân tích thành nhân tử
a)x8 +14x4 +1 b) x8 +98x4 +1
Thích thì bài đây
Lời giải:
$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$
$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
Bài 2:
1) \(x^2-4x+4=\left(x-2\right)^2\)
2) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
3) \(1-8x^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)
4) \(\left(x-y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)
6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
e)x3-4x+14x(x-2)=0
⇔ x(x2-4)+14x(x-2)=0
⇔ x(x-2)(x+2)+14x(x-2)=0
⇔ (x-2)(x2+2x+14x)=0
⇔ x(x-2)(x+16)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)
g)x2(x+1)-x(x+1)+x(x-1)=0
⇔ (x+1)(x2-x)+x(x-1)=0
⇔ x(x+1)(x-1)+x(x-1)=0
⇔ x(x-1)(x+2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a,\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)=\sqrt{a}\left(\sqrt{b-1}\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right).\left(\sqrt{b}-1\right).\)
mầy phần này dễ mà mình gại đánh máy quá
những phần sau sử dụng hằng đẳng thức nhé
T ms học lp 8 thôi mà . AHuhu =[[