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Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung
(3x + 2)^2 + (3x - 2)^2 - 2(9x^2 - 4)
\(=\left(3x+2\right)^2-2\left(3x+2\right)\left(3x-2\right)+\left(3x-2\right)^2\)
\(=\left(3x+2-\left(3x-2\right)\right)^2\)
\(=\left(3x+2-3x+2\right)^2\)
\(=4^2\)
\(=16\)
\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)
\(=\left(3x+2\right)^2+\left(3x-2\right)^2-2.\left(3x-2\right)\left(3x+2\right)\)
\(=\left(3x+2-3x+2\right)^2\)
\(=4^2=16\)
1.a) (3x+1)2-4(x-2)2= (3x+1)2-[2(x-2)]2=[(3x+1)-2(x-2)][(3x+1)+2(x-2)]=(x+3)(5x-1)
b) (a2+b2-5)2-4(ab+2)2= (a2+b2-5)2-[2(ab+2)]2 = (a2+b2-5-2ab-4)(a2+b2-5+2ab+4)=[(a-b)2-9][(a+b)2-1]
2. 3x2+9x-30=3x2-6x+15x-30=3x(x-2)+15(x-2)=3(x+5)(x-2)
b. x3-5x2-14x=x3+2x2-7x2-14x=x2(x+2)-7x(x+2)=(x2-7x)(x+2)
a) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+1\right)^2-\left[2.\left(x-2\right)\right]^2\)
\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left[3x+1-2x+4\right].\left[3x+1+2x-4\right]\)
\(=\left(x+5\right)\left(5x-3\right)\)
b) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left[2.\left(ab+2\right)\right]^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-9\right].\left[\left(a+b\right)^2-1\right]\)
\(=\left[\left(a-b-3\right)\left(a-b+3\right)\right].\left[\left(a+b-1\right)\left(a+b+1\right)\right]\)
a) \(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2-2x+5x-10\right)\)
\(=3.\left[x\left(x-2\right)+5.\left(x-2\right)\right]\)
\(=3.\left[\left(x+5\right)\left(x-2\right)\right]\)
b) \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2+2x-7x-14\right)\)
\(=x.\left[x\left(x+2\right)-7.\left(x+2\right)\right]\)
\(=x.\left[\left(x-7\right)\left(x+2\right)\right]\)
a)4a2b4-c4d2=(2ab2)2-(c2d)2=(2ab2-c2d)(2ab2+c2d)
b) (a+b)3-(a-b)3== 2a( a² + 2ab + b² - a² + b² + a² - 2ab + b² )
= 2a( a² + 3b²)
c)(6x-1)2-(3x+2)=36x2-12x+1-3x-2=36x2-15x-1=(6x)2-2.6x.\(\frac{15}{12}\)+\(\left(\frac{15}{12}\right)^2\)-\(\frac{41}{16}\)
=(6x-\(\frac{5}{4}\))2-\(\sqrt{\frac{41}{4}}^2\)=\(\left(6x-\frac{5}{4}-\sqrt{\frac{41}{4}}\right)\left(6x-\frac{5}{4}+\sqrt{\frac{41}{4}}\right)\)
\(3x^2\left(x-2\right)+6x\left(2-y\right)\)(1)
Nhân tử chung: \(3x\)
\(\Rightarrow\left(1\right)=3x\left[x\left(x-2\right)+2\left(2-y\right)\right]\)
\(=3x\left(x^2-2x+4-2y\right)\)
cái này có những biểu thức ko dùng được hđt hoặc đặt nhân tử chung =))) anh nghĩ đề em nên thay chữ và thành hoặc
1, \(15x^2+10xy=5x\left(3x+2y\right)\)
2, \(24x-18y+30=6\left(4x-3y+5\right)\)
3, \(2x\left(y-2009\right)+5y\left(y-2009\right)=\left(y-2009\right)\left(2x+5y\right)\)
4, \(35x\left(y-8\right)-14y\left(8-y\right)=\left(y-8\right)\left(35x+14y\right)=7\left(5x+2y\right)\left(y-8\right)\)
5, \(x^2+14x+49=x^2+2.7x+7^2=\left(x+7\right)^2\)
6, \(9x^2-4=\left(3x-2\right)\left(3x+2\right)\)
Trả lời:
1, 15x2 + 10xy = 5x ( 3x + 2y )
2, 24x - 18y + 30 = 6 ( 4x - 3y + 5 )
3, 2x ( y - 2009 ) + 5y ( y - 2009 ) = ( y - 2009 )( 2x + 5y )
4, 35x ( y - 8 ) - 14y ( 8 - y ) = 35x ( y - 8 ) + 14y ( y - 8 ) = ( y - 8 )( 35x + 14y ) = 7 ( y - 8 )( 5x + 2y )
5, x2 + 14x + 49 = ( x + 7 )2
6, 9x2 - 4 = ( 3x - 2 )( 3x + 2 )
45 + x3 - 5x2 - 9x
= (x3 - 5x2) - (9x - 45)
= x2(x - 5) - 9(x - 5)
= (x - 5)(x2 - 9)
= (x - 5)(x - 3)(x + 3)
TL:
\(45+x^3-5x^2-9x\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x+3\right)\left(x-3\right)\left(x-5\right)\)
Bài làm:
\(8x^3+4x^2-9x+30=8x^3+16x^2-12x^2-24x+15x+30\)
\(=8x^2\left(x+2\right)-12x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(8x^2-12x+15\right)\)
\(8x^3+4x^2-9x+30\)
\(=8x^3+16x^2-12x^2-24x+15x+30\)
\(=8x^2\cdot\left(x+2\right)-12x\cdot\left(x+2\right)+15x\cdot\left(x+2\right)\)
\(=\left(x+2\right)\cdot\left(8x^2-12x+15\right)\)
\(x^3+6x^2+9x\)
\(=x\left(x^2+6x+9\right)\)
\(=x\left(x^2+2.x.3+3^2\right)\)
\(=x\left(x+3\right)^2\)
x^3 +6x^2 +9x =x(x^2+ 6x +9)
=x(x+3)^2