\(D=\frac{9x^2+6x+1}{3x+1}\left(x\ne\frac{-1}{3}\right)\)

\(\Leftrightarrow D=\frac{\left(3x+1\right)^2}{3x+1}=3x+1\)

thay x=-4(tm) vào biểu thức D ta có: D=3.(-4)+1=-12+1=-11

vậy D=-11 với x=-4

\(18:\left(-567\right)=-4:\left(9x\right)\)

\(-4:\left(9x\right)=\frac{-2}{63}\)

\(9x=-4:\left(\frac{-2}{63}\right)\)

\(9x=126\)

\(x=14\)

HT

Đầu bài là gì vậy?

a: \(12x^3-9x^2+3x\)

\(=3x\cdot4x^2-3x\cdot3x+3x\cdot1\)

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

b: Sửa đề: \(x^2-y^2+6x+9\)

\(=\left(x^2+6x+9\right)-y^2\)

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

(2x^2-3x+1)(2x^2+5x+1)=9x^2

<=> (2x^2+5x+1- 8x)(2x^2 +5x+1)=9x^2

<=> (2x^2+5x+1)^2 -8x(2x^2+5x+1)=9x^2

<=>  (2x^2+5x+1)^2 -2*(4x)*(2x^2+5x+1)=9x^2

<=>  (2x^2+5x+1)^2 -2*(4x)*(2x^2+5x+1)+(4x)^2=9x^2+16x^2

<=> (2x^2+5x+1 - 4x)^2=25x^2

<=> (2x^2+x+1)^2=25x^2

<=> (2x^2+x+1)^2 - 25x^2 =0

<=>(2x^2+x+1-5x)(2x^2+x+1+5x)=0

<=>(2x^2-4x+1)(2x^2+6x+1)=0

<=> (2x^2-4x+1)=0 => 2( x^2 - 2x + 1/2)=0

                                <=> x^2-2x +1/2 =0

                                <=> (x^2-2x+1) -1/2 =0

                                <=> (x-1)^2 =1/2     =>  x-1 =căn(1/2)  => x=căn(1/2)+1

                                                              => x-1=-(căn(1/2)) => x=- (căn(1/2)) +1

Hoặc  2x^2 +6x +1=0 

         <=> x^2 + 3x +1/2 =0                

         <=> (x^2 + 2*(1.5)x + (1.5)^2) -(1.5)^2+1/2 =0

         <=> (x+1.5)^2 - 7/4 =0

         <=> (x+1.5)^2 = 7/4    =>        x+1.5 = căn(7/4) => x=căn(7/4) -1.5

                                           =>      x+1.5 =- căn(7/4) => x=-căn(7/4) -1.5

nhớ thanks bạn (+_+)