Ta có: \(\left\{{}\begin{matrix}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{matrix}\right.\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

\(\Rightarrow x=y=z=-1\)(do \(\left(x+1\right)^2,\left(y+1\right)^2,\left(z+1\right)^2\ge0\forall x,y,z\))

a) \(A=x^{2020}+y^{2020}+z^{2020}=\left(-1\right)^{2020}+\left(-1\right)^{2020}+\left(-1\right)^{2020}=1+1+1=3\)

b) \(B=\dfrac{1}{x^{2020}}+\dfrac{1}{y^{2020}}+\dfrac{1}{z^{2020}}=\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}=3\)

3x^2+3y^2+4xy-2x+2y+2=0

=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0

=>x=1 và y=-1

M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1

Ta có : \(x^2+2y+1=0;y^2+2z+1=0;z^2+2x+1=0\)

\(\Rightarrow x^2+2y+1=y^2+2z+1=z^2+2x+1\)

\(\Rightarrow x^2+2y+1-y^2-2z-1-z^2-2x-1=0\)

\(\Rightarrow\left(x^2-2x+1\right)-\left(y^2-2y+1\right)-\left(z^2+2z+1\right)=0\)

\(\Rightarrow\left(x-1\right)^2-\left(y-1\right)^2-\left(z+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-1\right)^2=0\\\left(z+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\y-1=0\\z+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=1\\z=-1\end{cases}}\)

Thay \(x=1;y=1;z=-1\)vào A ta có :

\(A=1^{2015}+1^{2016}+\left(-1\right)^{2017}=1+1-1=1\)

Vậy A = 1

Từ \(\hept{\begin{cases}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{cases}}\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\left(1\right)\)

Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{cases}\left(2\right)}\)

Từ \(\left(1\right)\)và \(\left(2\right)\):

\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+1=0\\y+1=0\\z+1=0\end{cases}}\)

\(\Rightarrow x=y=z=-1\)

\(\Rightarrow A=\left(-1\right)^{2015}+\left(-1\right)^{2016}+\left(-1\right)^{2017}=-1+1-1=-1\)

Vậy \(A=-1\)

Ta có\(\hept{\begin{cases}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{cases}}\)

\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0+0+0\)

\(\Leftrightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

Ta thấy \(\hept{\begin{cases}\left(x+1\right)^2\\\left(y+1\right)^2\\\left(z+1\right)^2\end{cases}\ge0\forall x,y,z}\)

Dấu"=" xảy ra khi \(\left(x+1\right)^2=\left(y+1\right)^2=\left(z+1\right)^2=0\)

\(\Rightarrow x=y=z=-1\)

Vậy giá trị của biểu thức A là:

\(A=\left(-1\right)^{2015}+\left(-1\right)^{2016}+\left(-1\right)^{2017}=-1\)

Vậy A=-1

Em không chắc đâu nha!

Từ đề bài suy ra \(0\le x;y;z\le1\Rightarrow x\left(1-x\right)\ge0\Rightarrow x\ge x^2\)

Tương tự với  y với z.Ta có:

\(P=\sqrt{x^2+x^2+x+1}+\sqrt{y^2+y^2+y+1}+\sqrt{z^2+z^2+z+1}\)

\(\le\sqrt{x^2+2x+1}+\sqrt{y^2+2y+1}+\sqrt{z^2+2z+1}\)

\(=\sqrt{\left(x+1\right)^2}+\sqrt{\left(y+1\right)^2}+\sqrt{\left(z+1\right)^2}\)

\(=\left|x+1\right|+\left|y+1\right|+\left|z+1\right|\)

\(=\left(x+y+z\right)+3=1+3=4\)

Dấu "=" xảy ra khi (x;y;z) = (0;0;1) và các hoán vị của nó.

Vậy....

Em sai chỗ nào xin các anh/ chị chỉ rõ ra giúp ạ, chứ tk sai mà không góp ý thế em cũng không biết đường nào mà tránh cái lỗi sai tương tự đâu ạ! Em cảm ơn.

(x-1)^2+|2y-3|=0

=>x-1=0 và 2y-3=0

=>x=1 và y=1,5

B=4*1^20+5*1^2*1,5-6*1,5+2

=4+7,5-9+2

=4,5

2x² + 2y² + 3xy - x + y + 1 = 0

2x² + 2y² + 4xy - xy - x + y + 1 = 0

(2x² + 2y² + 4xy) + (-xy - x) + (y + 1) = 0

2(x + y)² - x(y + 1) + (y + 1) = 0

2(x + y)² + (y + 1)(1 - x) = 0

Do (x + y)² \(\ge0\)

\(\Rightarrow\) 2(x + y)² \(\ge0\)

\(\Rightarrow\) 2(x + y)² + (y + 1)(1 - x) = 0 \(\Leftrightarrow\) (y + 1)(1 - x) = 0

\(\Rightarrow y+1=0;1-x=0\)

*) y + 1 = 0

y = -1

*) 1 - x = 0

x = 1

Với x = 1; y = -1, ta có:

B = [1 + (-1)]²⁰¹⁸ + (1 - 2)²⁰¹⁸ + (-1 - 1)²⁰¹⁸

= 1 + 2²⁰¹⁸