\(\left(x+y\right)^2=\left(x+1\right)\left(x+2\right)\Leftrightarrow x^2+2xy+y^2=x^2+3x+2\)

\(\Leftrightarrow2xy-3x-2+y^2=0\)

\(2x\left(8x-1\right)^2\left(4x-1\right)-9=512x^4-256x^3+40x^2-2x-9\)

\(=\left(2x-1\right)\left(4x+1\right)\left(64x^2-16x+9\right)\)

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

\(\Leftrightarrow x^2-x^2+2xy+3x+y^2-2=0\)

\(\Leftrightarrow2xy+3x+y^2-2=0\)

P/s : chả hiểu đề bài :))

Vì AB // CD nên \(\hept{\begin{cases}\widehat{A}+\widehat{D}=180^0\\\widehat{B}+\widehat{C}=180^0\end{cases}}\)(định lí hình thang)

Mà \(\widehat{A}=5\widehat{D}\)=> \(\widehat{5D}+\widehat{D}=180^0\)=> \(6\widehat{D}=180^0\)=> \(\widehat{D}=30^0\)(1)

Thay (1) vào \(\widehat{A}=5\widehat{D}\)ta có :

\(\widehat{A}=5\cdot30^0=150^0\)

Lại có : \(\widehat{B}=4\widehat{C}\)

=> \(4\widehat{C}+\widehat{C}=180^0\)

=> \(5\widehat{C}=180^0\)

=> \(\widehat{C}=36^0\)(2)

Thay (2) vào \(\widehat{B}=4\widehat{C}\)ta có :

=> \(\widehat{B}=4\cdot36^0=144^0\)

Vậy : ^A = 150⁰ , ^B = 144⁰ , ^C = 36⁰ , ^D = 30⁰

a) 2( x - 1 )² + ( x + 3 )² = 3( x - 2 )( x + 1 )

<=> 2( x² - 2x + 1 ) + x² + 6x + 9 = 3( x² - x - 2 )

<=> 2x² - 4x + 2 + x² + 6x + 9 = 3x² - 3x - 6

<=> 2x² - 4x + x² + 6x - 3x² + 3x = -6 - 2 - 9

<=> 5x = -17

<=> x = -17/5

b) ( x - 1 )² - 2( x - 3 ) = ( x + 1 )²

<=> x² - 2x + 1 - 2x + 6 = x² + 2x + 1

<=> x² - 2x - 2x - x² - 2x = 1 - 1 - 6

<=> -6x = -6

<=> x = 1 

c) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 6( x + 1 )² + 3x² = -33

<=> x³ - 9x² + 27x - 27 - ( x³ - 3³ ) + 6( x² + 2x + 1 ) + 3x² = -33

<=> x³ - 9x² + 27x - 27 - x³ + 27 + 6x² + 12x + 6 + 3x² = -33

<=> x³ - 9x² + 27x - x³ + 6x² + 12x + 3x² = -33 - 27 + 27 - 6

<=> 39x = -39

<=> x = -1

a) Đặt \(a=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x+3=a+4\\x-2=a-1\\x+1=a+2\end{cases}}\)

    Ta có: \(2a^2+\left(a+4\right)^2=3.\left(a-1\right)\left(a+2\right)\)

        \(\Leftrightarrow2a^2+a^2+4a+4=3.\left(a^2+a-2\right)\)

        \(\Leftrightarrow3a^2+4a+4=3a^2+3a-6\)

        \(\Leftrightarrow a=-10\)

         \(\Rightarrow x-1=-10\)

        \(\Leftrightarrow x=-9\)

Vậy \(S=\left\{-9\right\}\)

b)  Đặt \(b=x-1\)\(\Rightarrow\)\(\hept{\begin{cases}x-3=b-2\\x+1=b+2\end{cases}}\)

     Ta có: \(b^2-2.\left(b-2\right)=\left(b+2\right)^2\)

         \(\Leftrightarrow b^2-2b+4=b^2+4b+4\)

         \(\Leftrightarrow-6b=0\) 

         \(\Leftrightarrow b=0\)

          \(\Rightarrow x-1=0\)

         \(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

c) Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

        \(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+6\left(x^2+2x+1\right)+3x^2+33=0\)

        \(\Leftrightarrow6x^2+12x+6+3x^2+33=0\)

       \(\Leftrightarrow9x^2+12x+39=0\)

       \(\Leftrightarrow\left(9x^2+12x+4\right)+35=0\)

       \(\Leftrightarrow\left(3x+2\right)^2+35=0\)

   Vì \(\left(3x+2\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(3x+2\right)^2+35\ge35>0\forall x\)

         mà \(\left(3x+2\right)^2+35=0\)

   \(\Rightarrow\)\(\left(3x+2\right)^2+35=0\)vô nghiệm

Vậy \(S=\varnothing\)

a. \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)

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

\(\Leftrightarrow-x^3+5x^2-5x=0\)

\(\Leftrightarrow-x\left(x^2-5x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)

a) \(x\left(x-2\right)-x\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-2-x^2+4x-3\right)=0\)

\(\Leftrightarrow x\left(-x^2+5x-5\right)=0\)

\(\Leftrightarrow x\left(x-\frac{5+\sqrt{5}}{2}\right)\left(x-\frac{5-\sqrt{5}}{2}\right)=0\)

=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)

b) \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)

\(\Leftrightarrow-12=0\left(vn\right)\)

c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)

\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)

\(\Leftrightarrow2x=15\)

\(\Rightarrow x=\frac{15}{2}\)

Vì \(\left(x^2-2^2\right)^2\ge0\Rightarrow\left(x^2-2^2\right)^2+2010\ge2010\)

                                    \(\Leftrightarrow\frac{\left(x^2-2^2\right)+2010}{-2009}\le\frac{2010}{-2009}\)

Vậy Dmax=-2010/2009, dấu = xảy ra khi và chỉ khi \(x^2-2^2=0\Leftrightarrow x=\pm2\)

a) VT = x³ + 3x²y + 3xy² + y³ + x³ - 3x²y + 3xy² - y³

          = 2x³ + 6xy²

          = 2x( x² + 3y² ) = VP

=> đpcm

b) VT = x³ + 3x²y + 3xy² + y³ - ( x³ - 3x²y + 3xy² - y³ )

          = x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³

          = 3x²y + 2y³

          = 2y( 3x² + y² ) = VP

=> đpcm

a)

 \(VT=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

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

b)

\(VT=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

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

Ta có: \(D=\frac{\left(x^2-2^2\right)+2010}{-2009}=\frac{x^2+2006}{-2009}=\frac{3-x^2}{2009}-1\)

Để D đạt GTLN => \(\frac{3-x^2}{2009}\) đạt GTLN, mà \(3-x^2\le3\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x=0\)

Vậy Max(D) = \(-\frac{2006}{2009}\) khi x = 0

mình ghi sai đề nha

Ta có :

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

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

\(\RightarrowĐPCM\)

VT = x³ + y³ ( HĐT số 6 )

= x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy²

= ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² )

= ( x + y )³ - 3xy( x + y ) = VP

=> đpcm