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bằng phương pháp nào zậy bn????
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a,\(6x^3y^2-9x^2y^3+1^2x^2y^2\)
\(=x^2y^2\left(6x-9y+1\right)\)
b,\(2x\left(x-1\right)+3\left(1-x\right)\)
\(=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)\)
a,
\(6x^3y^2-9x^2y^3+1\cdot x^2\cdot y^2\)
\(=x^2y^2\left(6x-9y+1\right)\)
b,
\(2x\left(x-1\right)+3\left(1-x\right)\)
\(=2x\left(x-1\right)+3\cdot-1\left(x-1\right)\)
\(=2x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x-3\right)\left(x-1\right)\)
A=x^4+6x^3+7x^2–6x+1=x^4+(6x^3–2x^2)+(9x^2–6x+1)
= x^4+2x^2(3x–1)+(3x–1)^2 =(x^2+3x–1)^2
chỉnh lại tí
Đặt P(x)=x4+6x3+7x2- 6x+1
Đặt y=x2-1
=>y2=x4-2x2+1
P(x)=x4-2x2+1+6x3-6x+9x2
=(x2-1)2+6x(x2-1)+9x2
Q(y)=y2+6xy+9x2
=(y+3x)2
P(x)=(x2-1+3x)2
1a/ x3+x2+x+1=0
x2(x+1).(x+1)=0
=> x2(x+1)=0 x =1
hoặc =>[
x+1=0 x=-1
b/(x+2)2=x+2
x2+2.x.2+22 =x+2
x+x+4x+4=x+2
6x+4=x+2
....
c/(x+1)(6x2+2x)+(x-1)(6x2+2x)=0
x2-12 + (6x2+2x)2=0
=> x2-1 = 0 x=1
hoặc => [
(6x2+2x)2=0 x= 0
a) \(6x^2-x-1\)
\(=6x^2-3x+2x-1\)
\(=3x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\)
mk chỉnh đề
\(x^2-x-2=x^2-2x+x-2=x\left(x-2\right)+\left(x-2\right)=\left(x+1\right)\left(x-2\right)\)
\(x^2+6x+7=x^2+6x+9-2=\left(x+3\right)^2-2=\left(x+3-\sqrt{2}\right)\left(x+3+\sqrt{2}\right)\)
\(1;x^2-x-2\)
\(=x^2-2x+x-2\)
\(=x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
\(2,x^2+6x-7\)
\(=x^2-x+7x-7\)
\(=x\left(x-1\right)+7\left(x-1\right)=\left(x+7\right)\left(x-1\right)\)
\(x^3+6x^2-13x-42\)
\(x^3+6x^2-13x-42\)
\(=\left(x+7\right)\left(x-3\right)\left(x+2\right)\)
b, \(2x^3-x^2+3x+6\)
\(=2x^3+2x^2-3x^2-3x+6x+6\)
\(=2x^2\left(x+1\right)-3x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2-3x+6\right)\)
a,\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left(x^2-3x+8x-24\right)\)
\(=\left(x-1\right)\left(x\left(x-3\right)+8\left(x-3\right)\right)\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)