Phân tích đa thức thành nhân tử
a, √x^3 - 2√x - x
b, 4a - 4√a - 1
c, x^2 - √x + x - 1
d, x + √x - √xy - 2y + √y
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b) Ta có: \(x^3-x^2y-xy^2+y^3\)
\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)^2\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
a: \(=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b: \(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left(x^2y^2-9\right)\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
c: \(=x^2-8x+x-8\)
\(=x\left(x-8\right)+\left(x-8\right)\)
\(=\left(x-8\right)\left(x+1\right)\)
\(a,xy+y^2-x-y\)
\(=\left(xy+y^2\right)-\left(x+y\right)\)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(y-1\right)\)
\(---\)
\(b,\left(x^2y^2-8\right)^2-1\)
\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)
\(=\left[\left(xy\right)^2-9\right]\left(x^2y^2-7\right)\)
\(=\left(xy-3\right)\left(xy+3\right)\left(x^2y^2-7\right)\)
\(---\)
\(c,x^2-7x-8\)
\(=x^2+x-8x-8\)
\(=\left(x^2+x\right)-\left(8x+8\right)\)
\(=x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x-8\right)\)
\(Toru\)
b: =x^3+2x^2-x^2+4
=x^2(x+2)-(x+2)(x-2)
=(x+2)(x^2-x+2)
c: =x^3-2x^2+x^2-4
=x^2(x-2)+(x-2)(x+2)
=(x-2)(x^2+x+2)
d: =(x-y)(x+y)-7(x+y)
=(x+y)(x-y-7)
a: =64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2+y^2-4xy)(8x^2+y^2+4xy)
b: =x^8+2x^4+1-x^4
=(x^4+1)^2-x^4
=(x^4-x^2+1)(x^4+x^2+1)
=(x^4-x^2+1)(x^4+2x^2+1-x^2)
=(x^4-x^2+1)(x^2+1-x)(x^2+x+1)
c: =(x+1)(x^2-x+1)+2x(x+1)
=(x+1)(x^2-x+1+2x)
=(x+1)(x^2+x+1)
d: =(x^2-1)(x^2+1)-2x(x^2-1)
=(x^2-1)(x^2-2x+1)
=(x-1)^2*(x-1)(x+1)
=(x+1)(x-1)^3
\(a.25^2-4a^2+12ab-9b^2\\ =25^2-\left(4a^2+12ab-9b^2\right)\\ =25^2-\left(2a-3b\right)^2\\ =\left(25-2a+3b\right)\left(25+2a-3b\right)\\ b.x^3+x^2y-xy^2-y^3\\ =x^2\left(x+y\right)-y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-y^2\right)\\ =\left(x+y\right)\left(x+y\right)\left(x-y\right)\\ =\left(x+y\right)^2\left(x-y\right)\)
a: Ta có: \(25x^2-4a^2+12ab-9b^2\)
\(=25x^2-\left(2a-3b\right)^2\)
\(=\left(5x-2a+3b\right)\left(5x+2a-3b\right)\)
b: Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
a.
\(2x^3-x^2y+x^2+y^2-2xy-y=0\)
\(\Leftrightarrow x^2\left(2x-y+1\right)-y\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(2x-y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-y=0\\2x-y+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x^2\\y=2x+1\end{matrix}\right.\)
Thế vào pt đầu:
\(\left[{}\begin{matrix}x^3+x-2=0\\x\left(2x+1\right)+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x^2+x+2\right)=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(x^2-2xy+x=-y\)
Thế vào \(y^2\) ở pt dưới:
\(x^2\left(x^2-4y+3\right)+\left(x^2-2xy+x\right)^2=0\)
\(\Leftrightarrow x^2\left(x^2-4y+3\right)+x^2\left(x-2y+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\Rightarrow y=0\\x^2-4y+3+\left(x-2y+1\right)^2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)
\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)
\(\Leftrightarrow-2y+4y^2-8y+4=0\)
\(\Leftrightarrow...\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a: =2(x-2)+y(x-2)
=(x-2)(2+y)
b: \(=\left(x+y\right)^2-4=\left(x+y+2\right)\left(x+y-2\right)\)
c: =(x-7)(x+2)
a.
2x - 4 + xy - 2y
= 2(x-2) +y(x-2)
= (x-2)(y+2)
c.
x^2 - 5x - 14
= x^2 + 2x - 7x - 14
= x(x+2) - 7(x+2)
= (x-7)(x+2)
\(a,=4x^3\left(x+1\right)-x\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\\ =x\left(2x-1\right)\left(2x+1\right)\left(x+1\right)\\ b,=\left(a-1\right)^2-\left(b-c\right)^2\\ =\left(a-1-b+c\right)\left(a-1+b-c\right)\\ c,=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\\ =\left(x^2-9x+17\right)^2-9-72\\ =\left(x^2-9x+17\right)^2-81=\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)