Phân tích đa thức thành nhân tử: a) \(4ab+a^2-3a-12b\)
b) \(x^3+3x^2+3x+1-27y^3\)
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Câu 1:
a: \(=6x^2+5x+1-6x^2-6x+x+1=2\)
b: \(=a^3+1-a^2+1=a^3-a^2+2\)
\(x^3+3x^2+3x+1-27y^3\)
\(=\left(x+1\right)^3-27y^3\)
\(=\left(x+1-3y\right)\left(x^2+3xy+9y^2\right)\)
\(1,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\\ 2,4x^2-4x+1-25y^2=\left(2x-1\right)^2-\left(5y\right)^2=\left(2x-5y-1\right)\left(2x+5y-1\right)\\ c,x^2-3x-10=\left(x^2-5x\right)+\left(2x-10\right)=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+2\right)\)
1
a,\(\left(2x+1\right)\left(3x+1\right)-\left(6x-1\right)\left(x+1\right)\)
=\(6x^2+2x+3x+1-\left(6x^2+6x-x-1\right)\)
\(=6x^2+5x+1-6x^2-6x+x+1\)
\(=2\)
c,\(\left(a+1\right)\left(a^2-a+1\right)+\left(a+1\right)\left(a-1\right)\)
\(=\left(a^3+1\right)+\left(a^2-1\right)\)
\(=a^3+1+a^2-1\)
\(=a^3+a^2\)
2,
a,\(4ab+a^2-3a-12b\)
\(=\left(4ab-12b\right)+\left(a^2-3a\right)\)
\(=4b\left(a-3\right)+a\left(a-3\right)\)
\(=\left(4b+a\right)\left(a-3\right)\)
b,\(x^3+3x^2+3x+1-27y^3\)
\(=\left(x+1\right)^3-\left(3y\right)^3\)
\(=\left(x+1-3y\right)\left[\left(x+1\right)^2+\left(x+1\right).3y+\left(3y\right)^2\right]\)
\(=\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\)
4
a,\(2004^2-16\)
\(=2004^2-4^2\)
\(=\left(2004-4\right)\left(2004+4\right)\)
\(=2000.2008\)
\(=4016000\)
b,\(892^2+892.216+108^2\)
\(=\left(892+108\right)^2\)
\(=1000^2=1000000\)
c,\(10,2.9,8-9,8.0,2+10,2^2-10,2.0,2\)
\(=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)\)
\(=9,8.10+10,2.10\)
\(=98+102\)
\(=200\)
d,\(36^2+26^2-52.36\)
=\(\left(36-26\right)^2\)
\(=10^2=100\)
3)\(A=-x^2+2x-3\)
\(\Leftrightarrow A=-x^2+2x-1-2\)
\(\Leftrightarrow A=-\left(x^2-2x+1\right)-2\)
\(\Leftrightarrow A=-\left(x-1\right)^2-2\)
Vậy GTLN của A=-2 khi x=1
\(x^2+4x+4=\left(x+2\right)^2 \)
\(4x^2-4x+1=\left(2x-1\right)^2\)
\(c\left(x+1\right)-y\left(x+1\right)=\left(x+1\right)\left(c-y\right)\)
\(x^3-3x^2+3x-1+27y^3=\left(x-1\right)^3+27y^3=\left(x-1+3y\right)\left(x^2-2x+1-3xy+3y+9y^2\right)\)
`a)3x-3a+yx-ya`
`=3(x-a)+y(x-a)`
`=(x-a)(y+3)`
`b)x^2-9-4(x+3)`
`=(x-3)(x+3)-4(x+3)`
`=(x+3)(x-3-4)`
`=(x+3)(x-7)`
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
Phân tích đa thức thành nhân tử
27y2-9(x+y)2=\(9\left(3y^2-\left(x+y\right)^2\right)\)
=\(9\left(\sqrt{3}y+x+y\right)\left(\sqrt{3}y-x-y\right)\)
Rút gọn biểu thức
(2x4-x3+3x2): (-1/3x)
=\(\frac{2x^4-x^3+3x^2}{-\frac{1}{3x}}=3x^3\left(-2x^2+x-3\right)\)
a) \(4ab+a^2-3a-12b=a\left(4b+a\right)-3\left(4b+a\right)=\left(a-3\right)\left(4b+a\right)\)
b) \(x^3+3x^2+3x+1-27y^3=\left(x+1\right)^3-27y^3=\left(x+1-3y\right)\left[\left(x+1\right)^2+3y\left(x+1\right)+9y^2\right]=\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\)