\(T=\frac{\left(xy+z\right)\left(yz+x\right)\left(zx+y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=\frac{\left(xy+1-x-y\right)\left(yz+1-y-z\right)\left(zx+1-z-x\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=\frac{\left(x-1\right)\left(y-1\right)\left(y-1\right)\left(z-1\right)\left(z-1\right)\left(z-1\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=\frac{\left(x-1\right)^2\left(y-1\right)^2\left(z-1\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=\frac{\left(-y-z\right)^2\left(-x-z\right)^2\left(-x-y\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=\frac{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\)

\(T=1\). Vậy \(T\) không phụ thuộc vào \(x,y,z\).

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

\(M=x^3-6x^2+27x-27-x^3-3x^2-3x-1\)

\(M=-9x^2+24x-28\)

\(M=-\left(9x^2-24x+16\right)-12\)

\(M=-\left(3x-4\right)^2-12\)

Mà \(\left(3x-4\right)^2\ge0\Leftrightarrow-\left(3x-4\right)^2\le0\Leftrightarrow-\left(3x-4\right)^2-12\le-12\Leftrightarrow M\le-12\)

Dấu "=" xảy ra khi \(3x-4=0\Leftrightarrow x=\frac{4}{3}\)

Vậy GTLN của M là -12 khi \(x=\frac{4}{3}\)

\(=\frac{xyy+xzz+xxy+yzz+xxz+yyz}{xy+yz+xz-3}\)

\(=\frac{xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)}{xy+yz+xz-3}\)

\(=\frac{xy\left(xyz-z\right)+yz\left(xyz-x\right)+xz\left(xyz-y\right)}{xy+yz+xz-3}\)

\(=\frac{xyxyz+yzxyz+xzxyz-3xyz}{xy+yz+xz-3}\)

\(=\frac{xyz\left(xy+yz+xz-3\right)}{xy+yz+xz-3}\)

\(=xyz\)

x² + 4x -y² = 1

=> x² + 4x - y² + 4 = 1 + 4 = 5 

=> (x² + 4x + 4) - y² = 5 

=> (x+2)² - y² = 5

=> (x+2-y)(x+2+y) = 5

Ta có:

1.5=5

mà x+2-y < x+2+y

=> \(\hept{\begin{cases}\text{x+2-y=1}\\\text{x+2+y}=5\end{cases}}\)=> \(\hept{\begin{cases}x-y=-1\\x+y=3\end{cases}}\)

Từ x-y = -1 => x = y - 1

Thay x = y - 1 vào x + y, ta có:

x + y = y - 1 + y = 3

=> 2y - 1 = 3

=> 2y = 4 => y=2

=> x = 2 - 1 = 2

Vậy x=2; y = 1 thì x² + 4x -y² = 1

Đáp án trong hình

2 - 1 = 1 chứ

a) ( x + 2 )² + (x + 3)² + (x + 4)⁴ = 2

Đặt x + 3 = t 

=>  PT trở thành  (t - 1)² + t³ + (t + 1)⁴ = 2

Ta có: (t - 1)² + t³ + (t + 1)⁴ = 2

<=> t² - 2t + 1 + t³ + ( t + 1)^2.. (t + 1)² =2

<=>  t² - 2t + 1 + t³  +  (t² + 2t + 1) . (t² + 2t + 1) = 2 

<=> t² - 2t + 1 + t³ + ( t⁴ + 2t³ + t² + 2t³ + 4t² + 2t + t² + 2t + 1) = 2

<=> t² - 2t + 1 + t³ + t⁴ + 4t³ + 6t² + 4t - 2 + 1 = 0 

<=> t⁴ + 5t³ +7t² + 2t = 0

<=> t . (t³ + 5t² + 7t + 2) =0

<=> t . ( t³ + 2t² + 3t² + 6t + t + 2) =0

<=> t . [ t² . ( t + 2) + 3t . (t + 2 ) +(t+2)] =0

<=> t.  (t + 2) (t² + 3t + 1) = 0 

<=> ( t + 2) ( t² + 3t + 1) = 0 

=> 2 th :

* t + 2 = 0 <=> t = -2 ( t/m)

* t² + 3t + 1 = 0 

<=>  t² + 3t + 1 = 0 

<=> t² + 2t + t + 1 =0

<=> ( t² + 2t + 1) + t = 0 

<=> (t + 1)² + t =0

Vì: (t + 1)² > 0 => (t + 1)² + t  > 0 ( ktm)

Vậy pt có nghiệm là S = {-2}

2Ta có :

x+(x-1)(x mũ 2 -2x+2)=0

Nên x=0 

bài toán nâng cao lớp 6 thì phải bởi vì em học lớp 6