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H
21 tháng 8 2021
Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
21 tháng 8 2021
1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)
2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)
4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
CC
2
SX
1
M
30 tháng 10 2015
\(4x^3-7x^2+3x\)
\(=4x^3-4x^2-3x^2+3x\)
\(=4x^2\left(x-1\right)-3x\left(x-1\right)\)
\(=\left(x-1\right)\left(4x^2-3x\right)=x\left(x-1\right)\left(4x-3\right)\)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-15\)
\(=\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)-15\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-15\)
\(=\left(x^2-5x+4\right)^2+2\left(x^2-5x+4\right)+1-16\)
\(=\left(x^2-5x+4+1\right)^2-4^2\)
\(=\left(x^2-4x+4+1-4\right)\left(x^2-4x+4+1+4\right)\)
\(=\left(x^2-4x+1\right)\left(x^2-4x+9\right)\)
PQ
1
1 tháng 11 2015
a) x3 - 7x - 6 = x3 + x2 - x2 - x - 6x - 6
= x2(x + 1) - x(x + 1) - 6(x + 1)
= (x + 1)(x2 - x - 6)
= (x + 1)(x2 + 2x - 3x - 6)
= (x + 1)[x(x + 2) - 3(x + 2)]
= (x + 1)(x + 2)(x - 3)
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
25 tháng 7 2021
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
25 tháng 7 2021
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
17 tháng 10 2021
\(a,=2x^2-6x+9x-27=\left(x-3\right)\left(2x+9\right)\\ b,=x^2-7x+\dfrac{49}{4}-\dfrac{73}{4}\\ =\left(x-\dfrac{7}{2}\right)^2-\dfrac{73}{4}=\left(x-\dfrac{7}{2}-\dfrac{\sqrt{73}}{2}\right)\left(x-\dfrac{7}{2}+\dfrac{\sqrt{73}}{2}\right)\\ c,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ d,=x^2-2x-8x+16=\left(x-2\right)\left(x-8\right)\\ e,=x^2-3x-5x+15=\left(x-3\right)\left(x-5\right)\\ g,=x^2+2x+4x+8=\left(x+2\right)\left(x+4\right)\)
a) 4x2 + 4x - 3x = 4x2 +x = x( 4x+1)
b) x2+7x+10= x2+2x+5x+10= x(x+2)+5(x+2)= (x+5)(x+2)
c) x2-x-12= x2 - 4x+3x-12= x(x-4)+3(x-4)=(x+3)(x-4)
d) x2+3x-18=x2+6x-3x-18= x(x+6)-3(x+6)=(x-3)(x+6)
\(x^2+7x+10\)
\(=x^2+2x+5x+10\)
\(=x\left(x+2\right)+5\left(x+2\right)\)
\(=\left(x+2\right)\left(x+5\right)\)