a) ( 4 - x )( x² - 7x + 1 ) - x² + 16 = 0

<=> ( 4 - x )( x² - 7x + 1 ) + ( 16 - x² ) = 0

<=> ( 4 - x )( x² - 7x + 1 ) + ( 4 - x )( 4 + x ) = 0

<=> ( 4 - x )( x² - 7x + 1 + 4 + x ) = 0

<=> ( 4 - x )( x² - 6x + 5 ) = 0

<=> ( 4 - x )( x - 2 )( x - 3 ) = 0

<=> x = 2 hoặc x = 3 hoặc x = 4

Vậy phương trình có tập nghiệm S = { 2 ; 3 ; 4 }

b) 4x³ + 7x² + 7x + 4 = 0

<=> ( 4x³ + 4 ) + ( 7x² + 7x ) = 0

<=> 4( x³ + 1 ) + 7x( x + 1 ) = 0

<=> 4( x + 1 )( x² - x + 1 ) + 7x( x + 1 ) = 0

<=> ( x + 1 )[ 4x² - 4x + 4 + 7x ) = 0

<=> ( x + 1 )( 4x² + 3x + 4 ) = 0

Vì 4x² + 3x + 4 > 0 => x + 1 = 0 <=> x = -1

Vậy phương trình có nghiệm x = -1

mơn bn

( 3x + 6 )( -x - 9 ) = ( 3x + 6 )( x - 3 )

<=> ( 3x + 6 )( -x - 9 ) - ( 3x + 6 )( x - 3 ) = 0

<=> ( 3x + 6 )( -x - 9 - x + 3 ) = 0

<=> ( 3x + 6 )( -2x - 6 ) = 0

<=> 3x + 6 = 0 hoặc -2x - 6 = 0

<=> x = -2 hoặc x = -3

Vậy phương trình có tập nghiệm S = { -2 ; -3 }

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

\(\Leftrightarrow3x+4=0\)

\(\Leftrightarrow3x=-4\)

\(\Leftrightarrow x=\frac{-4}{3}\)

Vậy phương trình \(\left(3x+4\right)^2=0\)có tập nghiệm là \(S=\left\{\frac{-4}{3}\right\}\)

Đặt \(x-2016=a\) khi đó PT trở thành:

\(PT\Leftrightarrow\frac{\left(a+1\right)^2+a\left(a+1\right)+a^2}{\left(a+1\right)^2-a\left(a+1\right)+a^2}=\frac{19}{49}\)

\(\Leftrightarrow\frac{a^2+2a+1+a^2+a+a^2}{a^2+2a+1-a^2-a+a^2}=\frac{19}{49}\)

\(\Leftrightarrow\frac{3a^2+3a+1}{a^2+a+1}=\frac{19}{49}\)

\(\Leftrightarrow147a^2+147a+49=19a^2+19a+19\)

\(\Leftrightarrow128a^2+128a+30=0\)

\(\Leftrightarrow64a^2+64a+15=0\)

\(\Leftrightarrow\left(64a^2+24a\right)+\left(40a+15\right)=0\)

\(\Leftrightarrow8a\left(8a+3\right)+5\left(8a+3\right)=0\)

\(\Leftrightarrow\left(8a+3\right)\left(8a+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{8}\\a=-\frac{5}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-2016=-\frac{3}{8}\\x-2016=-\frac{5}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{16125}{8}\\x=\frac{16123}{8}\end{cases}}\)

Vậy: ...

Ta có : \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=4\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\Rightarrow x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2=4\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\Rightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=4\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\Rightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x,y\\\left(y-z\right)^2\ge0\forall y,z\\\left(z-x\right)^2\ge0\forall z,x\end{cases}}\)\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\)

\(\Rightarrow x=y=z\left(đpcm\right)\)

 Biến đổi vế trái ta có

  \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)

\(=2x^2+2y^2+2z^2-2xy-2yz-2zx\)

Biến đổi vế phải ta có

  \(4\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(=4x^2+4y^2+4z^2-4xy-4yz-4zx\)

Theo bài

   \(2x^2+2y^2+2z^2-2xy-2yz-2zx=4x^2+4y^2+4z^2\)

                          \(-4xy-4yz-4zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

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

 Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}}\)

 Để \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow}x=y=z}\)