TH1:x,y,z=0

TH2:x=2\(\frac{3}{10}\)

y=3\(\frac{5}{6}\)

z=11\(\frac{1}{2}\)

giải ra cơ kết quả mik cx có mà hình như KQ sai rồi

à đúng rồi mà cách giải là sao v chỉ mik vs

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\) \(\Leftrightarrow x=y=z\)

Mà \(x+y+z=-3\Rightarrow x=y=z=-1\)

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

a)

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

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

b) 

\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=x^3+x^2y+x^2z+xy^2+y^3+y^2z+\)

\(+xz^2+yz^2+z^3-x^2y-xy^2-xyz-xyz-y^2z-yz^2-x^2z-xyz-xz^2=\)

\(=x^3+y^3+z^3-3xyz\)

Không mất tính tổng quát, giả sử \(x=mid\left\{x;y;z\right\}\)

\(\Rightarrow\left(x-y\right)\left(x-z\right)\le0\)

\(\Rightarrow x^2+yz\le xy+xz\)

\(\Rightarrow zx^2+yz^2\le xyz+xz^2\)

\(\Rightarrow P\le x^3+y^3+z^3+8\left(xy^2+xz^2+xyz\right)\)

\(\Rightarrow P\le x^3+y^3+z^3+3yz\left(y+z\right)+8\left(xy^2+xz^2+2xyz\right)\)

\(\Rightarrow P\le x^3+\left(y+z\right)^3+8x\left(y+z\right)^2\)

\(\Rightarrow P\le x^3+\left(4-x\right)^3+8x\left(4-x\right)^2\)

\(\Rightarrow P\le8x^3-52x^2+80x+64\)

Tới đây, đơn giản nhất là khảo sát hàm \(f\left(x\right)=8x^3-52x^2+80x+64\) trên \(\left[0;4\right]\)

(Nếu ko khảo sát hàm, ta có thể tách như sau, tất nhiên là dựa trên điểm rơi có được từ việc khảo sát hàm):

\(\Rightarrow P\le\left(8x^3-52x^2+80x-36\right)+100\)

\(\Rightarrow P\le4\left(x-1\right)^2\left(2x-9\right)+100\)

Do \(0\le x\le4\Rightarrow2x-9< 0\Rightarrow P\le100\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;3;0\right)\) và 1 vài bộ hoán vị của chúng

1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)

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

2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)

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

3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)

\(=5x\left(x-4\right)+2\left(x-4\right)=\left(x-4\right)\left(5x+2\right)\)

4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)

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

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

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

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

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

\(=\left(x+y\right)\left(xy+xz-yz-z^2\right)=\left(x+y\right).\left[x\left(y+z\right)-z\left(y+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(x-z\right)\)

1) 4x² - 7x - 2 = 4x² - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )

2) 4x² + 5x - 6 = 4x² - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )

3) 5x² - 18x - 8 = 5x² - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )

4) xy( x + y ) - yz( y + z ) + xz( x - z )

= x²y + xy² - y²z - yz² + xz( x - z )

= ( x²y - yz² ) + ( xy² - y²z ) + xz( x - z )

= y( x² - z² ) + y²( x - z ) + xz( x - z )

= y( x - z )( x + z ) + y²( x - z ) + xz( x - z )

= ( x - z )[ y( x + z ) + y² + xz ]

= ( x - z )( xy + yz + y² + xz )

= ( x - z )[ ( xy + y² ) + ( xz + yz ) ]

= ( x - z )[ y( x + y ) + z( x + y ) ]

= ( x - z )( x + y )( y + z )

5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )

\(4x^2-7x-2=\left(4x^2-8x\right)+\left(x-2\right)=4x\left(x-2\right)+\left(x-2\right)=\left(4x-1\right)\left(x-2\right)\)

\(=4x^2+8x-3x-6=4x\left(x+2\right)-3\left(x+2\right)=\left(4x-3\right)\left(x+2\right)\)

\(=5x^2-18x-8=5x^2-20x+2x-8=5x\left(x-4\right)+2\left(x-4\right)=\left(5x+2\right)\left(x-4\right)\)

\(5=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)