Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Sửa đề: \(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>\(\left\{{}\begin{matrix}2x+2y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(M=\left(x-y\right)^{2023}-\left(x-2\right)^{2024}+\left(y+1\right)^{2023}\)
\(=\left(1+1\right)^{2023}-\left(1-2\right)^{2024}+\left(-1+1\right)^{2023}\)
\(=2^{2023}-1\)
\(5x^2+5y^2+8xy-2x+2y+2=0\)
=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)
=>\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
=>x=1 và y=-1
\(M=\left(1-1\right)^{2023}+\left(1-2\right)^{2024}+\left(-1+1\right)^{2025}=1\)
\(x^2-xy-2022x+2023y-2024=0\\\Leftrightarrow (x^2-2023x)-(xy-2023y)+(x-2023)-1=0\\\Leftrightarrow x(x-2023)-y(x-2023)+(x-2023)=1\\\Leftrightarrow(x-2023)(x-y+1)=1\)
Vì \(x,y\) nguyên nên \(x-2023;x-y+1\) có giá trị nguyên
mà \(\left(x-2023\right)\left(x-y+1\right)=1\)
nên ta có các trường hợp xảy ra là:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2023=1\\x-y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2023=-1\\x-y+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=2024\left(tm\right)\\\left\{{}\begin{matrix}x=2022\\y=2024\end{matrix}\right.\left(tm\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2024;2024\right);\left(2022;2024\right)\).
\(\text{#}Toru\)
\(\sqrt{x^2+2024}=\sqrt{x^2+xy+yz+zx}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)
Tương tự: \(\sqrt{y^2+2024}\ge\sqrt{xy}+\sqrt{yz}\)
\(\sqrt{z^2+2024}\ge\sqrt{xz}+\sqrt{yz}\)
Cộng vế:
\(P\ge\dfrac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=2\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2024}{3}\)
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}+\dfrac{x+3}{2013}+\dfrac{x+4}{2012}+\dfrac{x+2024}{2}=0\)
\(\Leftrightarrow(\dfrac{x+1}{2015}+1)+(\dfrac{x+2}{2014}+1)+(\dfrac{x+3}{2013}+1)+(\dfrac{x+4}{2012}+1)+\dfrac{x+2024}{2}-4=0\)\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}+\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}+\dfrac{x+2016}{2}=0\)\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}\right)=0\)
Hiển nhiên: \(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2}>0\)
\(\Leftrightarrow x+2016=0\Leftrightarrow x=-2016\)
2(x - 2)²⁰²⁴ + 5|y - 1| = 0
Do (x - 2)²⁰²⁴ ≥ 0 với mọi x
|y - 1| ≥ 0 với mọi y
2(x - 2)²⁰²⁴ + 5|y - 1| = 0
⇔ x - 2 = 0 và y - 1 = 0
*) x - 2 = 0
x = 2
*) y - 1 = 0
y = 1
Vậy x = 2; y = 1