Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0

=> 6x² - 21x - (6x² + x - 90x - 15) - 2010 = 0

=> 6x² - 21x - 6x² + 89x + 15 - 2010 = 0

=> 68x - 1995 = 0

 ? 

b) 2x(x - 2012) - x + 2012 = 0

=> 2x(x - 2012) - (x - 2012) = 0

=> (x - 2012) (2x - 1) = 0

⇔[

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Vậy x = {2012;12 }

Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0

=> 6x² - 21x - (6x² + x - 90x - 15) - 2010 = 0

=> 6x² - 21x - 6x² + 89x + 15 - 2010 = 0

=> 68x - 1995 = 0

 ? 

b) 2x(x - 2012) - x + 2012 = 0

=> 2x(x - 2012) - (x - 2012) = 0

=> (x - 2012) (2x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-2012=0\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2012\\2x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2012\\x=\frac{1}{2}\end{cases}}\)

Vậy x = \(\left\{2012;\frac{1}{2}\right\}\)

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

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

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

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

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

c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)

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

d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)

a) ( x+2y)(x^2-2xy+4y^2)-8y^3+27=0

x^3+(2y)^3 - (2y)^3 + 3^3 = 0

x^3+3^3 = 0

(x+3)(x^2-3x+9)= 0

=> x+3=0 hoặc x^2 - 3x + 9 = 0

x+3 = 0 => x=-3

x^2-3x+9= 0=> x^2-2x.3/2+9/4-9/4+9=0

(x^2-2x.3/2+9/4)+(-9/4+9)=0

(x-3/2)^2 + 25/4=0

(x-3/2)^2 =-25/4

Vì (x-3/2)^2 >= 0 mà -25/4<0 nên k tìm đc x tỏa mãn đk đề bài

Vậy x=-3

b) x^3 + x^2 - 2x - 8 = 0

(x^3-8)+(x^2-2x)=0

(x-2)(x^2+2x+4)+x(x-2)=0

(x-2)(x^2+2x+4+x)=0

(x-2)(x^2+3x+4)=0

=>x-2=0 hoặc x^2+3x+4=0

x-2=0=>x=2

x^2+3x+4=0

x^2+2x.3/2+9/4-9/4+4=0

(x^2+2x.3/2+9/4)+(-9/4+4)=0

(x+3/2)^2+15/4=0

(x+3/2)^2=-15/4

Vì (x+3/2)^2>=0 mà-15/4<0 nên k tìm đc x thỏa mãn đk đề bài

Vậy x=2

Mình chỉ làm đc như này thui b thông cảm

Tick cho mình nha

Câu a :

\(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-8x^3+27=0\)

\(\Leftrightarrow\left(x^3+8y^3\right)-\left(8y^3-27\right)=0\)

\(\Leftrightarrow x^3+8y^3-8y^3-27=0\)

\(\Leftrightarrow x^3-27=0\)

\(\Rightarrow x=3\)

Câu b :
\(x^3+x^2-2x-8=0\)

\(\Leftrightarrow\left(x^3-8\right)+\left(x^2-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+3x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=2\)

\(x^3+8y^3+2xy^2+x^2y\)

\(=x^3+2x^2y-x^2y-2xy^2+4xy^2+8y^3\)

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

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

\(A=5x^2-3x-x^3+x^2+x^3-62x-10+3x\\ A=6x^2-62x-10\\ B=x^3+x^2+x-x^3-x^2-x+5=5\\ C=3x^2y-15xy^2+15xy^2-10y^3+10y^2-3x^2y-4=-4\)

b: Ta có: \(B=x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)

\(=x^3+x^2+x-x^3-x^2-x+5\)

=5

P=\(X^2+2Y^2-2XY+8X+8Y+2017\)

P=\(\dfrac{4X^2+8Y^2-8XY+32Y+32X+8068}{4}\)

P=\(\dfrac{(\sqrt{3}X)^2-2.\sqrt{3}X.\dfrac{4}{\sqrt{3}}Y+\left(\dfrac{4}{\sqrt{3}}Y\right)^2-\left(\dfrac{4}{\sqrt{3}}Y\right)^2+8Y^2+X^2+32X+32Y+8068}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+\dfrac{8}{3}Y^2+32X+32Y+8068}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+2.X.16+16^2+(\dfrac{2\sqrt{2}}{\sqrt{3}}Y)^2+2.\dfrac{2\sqrt{2}}{\sqrt{3}}Y.4\sqrt{6}+\left(4\sqrt{6}\right)^2+7716}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+\left(X+16\right)^2+\left(\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}\right)^2}{4}+1929\ge1929\forall X\in R\)

DẤU = XẢY RA \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y=0\\X+16=0\\\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}=0\end{matrix}\right.\)