Cứu với ;-;

Ta có: \(\hept{\begin{cases}\left(x-1\right)^{2008}=\left[\left(x-1\right)^{1004}\right]^2\ge0\\\left(y-2\right)^{2020}=\left[\left(y-2\right)^{1010}\right]^2\ge0\\\left(x+y-z\right)^{2022}=\left[\left(x+y-z\right)^{1011}\right]^2\ge0\end{cases}}\)

=> Tổng của 3 số dương =0 khi và chỉ khi cả 3 số đều bằng 0

=> \(\hept{\begin{cases}\left[\left(x-1\right)^{1004}\right]^2=0\\\left[\left(y-2\right)^{1010}\right]^2=0\\\left[\left(x+y-z\right)^{1011}\right]^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x-1=0\\y-2=0\\x+y-z=0\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)

Đáp số: x=1, y=2, z=3

Ta thấy: \(\hept{\begin{cases}\left(x-3\right)^{2020}\ge0\\\left(y-z\right)^{2022}\ge0\\\left|x-y-z\right|\ge0\end{cases}\left(\forall x,y,z\right)}\)

\(\Rightarrow\left(x-3\right)^{2020}+\left(y-z\right)^{2022}+\left|x-y-z\right|\ge0\left(\forall x,y,z\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-3\right)^{2020}=0\\\left(y-z\right)^{2022}=0\\\left|x-y-z\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=z\\y+z=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=z=\frac{3}{2}\end{cases}}\)

Vậy x = 3 và y = z = 3/2

Ta có : \(\hept{\begin{cases}\left(x-3\right)^{2020}\ge0\forall x\\\left(y-z\right)^{2022}\ge0\forall y;z\\\left|x-y-z\right|\ge0\forall x;y;z\end{cases}\Rightarrow}\left(x-3\right)^{2020}+\left(y-z\right)^{2022}+\left|x-y-z\right|\ge0\forall x;y;z\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-3=0\\y-z=0\\x-y-z=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=z\\x=y+z\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=1,5\\z=1,5\end{cases}}\)

Vậy x = 3 ; y = 1,5 ; z = 1,5 là giá trị cần tìm

hỏi khó thế anh zai

( x - 1 )²⁰¹⁸ + ( y + 3 )²⁰²⁰ + ( z - 5 )²⁰²² = 0

Ta thấy : ( x - 1 )²⁰¹⁸ \(\ge0\) ; ( y + 3 )²⁰²⁰ \(\ge0\) ; ( z - 5 )²⁰²² \(\ge0\)

\(\Rightarrow\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left(z-5\right)^{2022}\ge0\)

Theo đề,ta có : \(\left(x-1\right)^{2018}=\left(y+3\right)^{2020}=\left(z-5\right)^{2022}=0\)

+) \(\left(x-1\right)^{2018}=0\Rightarrow x-1=0\Rightarrow x=1\)

+) \(\left(y+3\right)^{2020}=0\Rightarrow y+3=0\Rightarrow y=-3\)

=) \(\left(z-5\right)^{2022}=0\Rightarrow z-5=0\Rightarrow z=5\)

Vậy : x = 1 ; y = -3 ; z = 5

\(\text{Ta có:}\)

\(\hept{\begin{cases}\left(x-1\right)^{2018}\ge0\\\left(y+3\right)^{2020}\ge0\\\left(z-5\right)^{2022}\ge0\end{cases}}\text{mà:}\left(x-1\right)^{2018}+\left(y-2\right)^{2020}+\left(z-3\right)^{2022}=0\text{ nên:}\)

\(\hept{\begin{cases}\left(x-1\right)^{2018}=0\\\left(y+3\right)^{2018}=0\\\left(z-5\right)^{2018}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-3\\z=5\end{cases}}\)

bạn tự kết luận

( x - 1 )²⁰¹⁸ + (y - 2 )²⁰²⁰+(z-3)²⁰²²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)

\(A=\dfrac{1}{9}\left(-x\right)^{2021}y^2z^3=\dfrac{1}{3}\left(-1\right)^{2021}.2^2.3^3=\dfrac{1}{3}.\left(-1\right).4.27=-36\)

Bài tập đâu rồi?

(Nó có hơi dài dòng)

Cho 3 số x,y,z thỏa mãn: x/2020=y/2021=z/2022.Chứng minh rằng: (x-z)^3 =

(x-z)^3= (2020 - 2022)^3 = -8

8(x-y)^2.(y-z)= 8(2020 - 2021)^2 . (2021 - 2022) = -8.

Vì (x-z)^3 = -8

 8(x-y)^2.(y-z) = -8

==> (x-z)^3 = 8(x-y)^2.(y-z)