a, 4x(x-5)-x(4x-7)=12

<=>4x²-20x-4x+7x=12

<=>-13x=12

<=>x=-12/13

b,(2-3x)x+6x(x-7)=4

<=>2x-3x²+6x²-42x=4

<=>3x²-40x=4

Đến đây thì mid

a, x= -0.99996

b, x= -0.286334219

c, x= -0.885584228

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

a: \(2y\left(x+2\right)-3x-6\)

\(=2y\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(2y-3\right)\)

b: \(3\left(x+4\right)-x^2-4x\)

\(=3\left(x+4\right)-\left(x^2+4x\right)\)

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(x+4\right)\left(3-x\right)\)

c: \(2\left(x+5\right)-x^2-4x\)

\(=2x+10-x^2-4x\)

\(=-x^2-2x+10\)

\(=-x^2-2x-1+11\)

\(=11-\left(x^2+2x+1\right)\)

\(=11-\left(x+1\right)^2\)

\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)

d: \(x^2+6x-3x-18\)

\(=\left(x^2+6x\right)-\left(3x+18\right)\)

\(=x\left(x+6\right)-3\left(x+6\right)\)

\(=\left(x+6\right)\left(x-3\right)\)

a)   \(=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)

\(=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x^3-2x^2+2x-1\right)\left(x-1\right)\)

\(=\left(x^3-x^2-x^2+x+x-1\right)\left(x-1\right)\)

\(=\left(x^2-x+1\right)\left(x-1\right)^2\)

c)

\(=6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6\)

\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)

\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)

\(=\left(6x^3+18x^2-x^2-3x-x-3\right)\left(x-2\right)\)

\(=\left(6x^2-x-1\right)\left(x+3\right)\left(x-2\right)\)

\(=\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\left(x-2\right)\)

b)

\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)

\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)

\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)

CÓ:   \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)

=>   \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)

=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.

Đầy đủ giúp em nhé

giup e với

\(a,=x\left(4x^2-1\right)=x\left(2x-1\right)\left(2x+1\right)\\ b,=2\left(3x-2\right)-x\left(3x-2\right)=\left(2-x\right)\left(3x-2\right)\\ c,=x\left(y-1\right)-\left(y-1\right)=\left(x-1\right)\left(y-1\right)\\ d,=\left(y+2\right)^2-4x^2=\left(y+2-2x\right)\left(y+2+2x\right)\\ e,=x^2-x-2x+2=\left(x-1\right)\left(x-2\right)\)

a) \(4x^3-x^2=x^2\left(4x-1\right)\)

b) \(6x-4+x\left(2-3x\right)=2\left(3x-2\right)-x\left(3x-2\right)=\left(2-x\right)\left(3x-2\right)\)

c) \(xy+1-x-y=\left(xy-x\right)-\left(y-1\right)=x\left(y-1\right)-\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)

d) \(y^2-4x^2+4y+4=\left(y^2+4y+4\right)-4x^2=\left(y+2\right)^2-\left(2x\right)^2=\left(y-2x+2\right)\left(y+2x+2\right)\)

e) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)