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Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
a: \(12x^3-6x^2+3x\)
\(=3x\cdot4x^2-3x\cdot2x+3x\cdot1\)
\(=3x\left(4x^2-2x+1\right)\)
b: \(\dfrac{2}{5}x^2+5x^3+x^2y\)
\(=x^2\cdot\dfrac{2}{5}+x^2\cdot5x+x^2\cdot y\)
\(=x^2\left(\dfrac{2}{5}+5x+y\right)\)
c: \(14x^2y-21xy^2+28x^2y^2\)
\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)
\(=7xy\left(2x-3y+4xy\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))
a) \(27x^3-8=\left(3x-2\right)\left(9x^2+6x+4\right)\)
b) \(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)
c) \(\left(2y-1\right)^1-4x^2+4x-1=\left(2y-1\right)^2-\left(2x-1\right)^2=\left(2y-1-2x+1\right)\left(2y-1+2x-1\right)\)
\(=\left(2y-2x\right)\left(2y+2x-2\right)=4\left(y-x\right)\left(y+x-1\right)\)
Bài làm:
a, 1-4x2
=1-(2x)2
=(1-2x).(1+2x)
b, 8-27x3
=23-(3x)3
=(2-3x).(4+6x+9x2)
Các câu còn lại bạn dùng hằng đẳng thức là phân tích được ra thôi
1 - 4x^2
= 1^2 - ( 2x )^2
= ( 1 - 2x ) ( 1 + 2x )
8 - 27x^ 3
= 2^3 - ( 3x )^3
= ( 2 - 3x ) [ 2^2 + 2 * 3x + ( 3x )^2 ]
= ( 2 - 3x ) ( 4 + 6x + 9x^2 )
= ( 2 - 3x ) ( 9x^2 + 6x + 4 )
27 + 27x + 9x^2 + x^3
= x^3 + 9x^2 + 27x + 27
= x^3 + 3x^2 + 6x^2 + 18x + 9x + 27
= x^2 ( x + 3 ) + 6x ( x + 3 ) + 9 ( x + 3 )
= ( x + 3 ) ( x^2 + 6x + 9 )
= ( x + 3 ) ( x + 3 )^2
= ( x + 3 )^3
x^2 + 4x - 5
= x^2 - x + 5x - 5
= x ( x - 1 ) + 5 ( x - 1 )
= ( x + 1 ) ( x - 5 )
\(\left(x-1\right)^2-25\)
\(=x^2-2x+1-25\)
\(=x^2-2x-24\)
\(=x^2-6x+4x-24\)
\(=x.\left(x-6\right)+4.\left(x-6\right)\)
\(=\left(x+4\right).\left(x-6\right)\)
a, \(1-2y+y^2=\left(y+1\right)^2=\left(y+1\right)\left(y+1\right)\)
b, \(\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
c, \(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)
d, \(8-27x^3=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
Bài 1:
\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)
Bài 2:
\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)