Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=0\Leftrightarrow2x^2+y^2+5+2xy-6x-4y=0\)
\(\Leftrightarrow x^2+y^2+4+2xy-4x-4y+x^2-2x+1=0\)
\(\Leftrightarrow\left(x+y-2\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-2=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(\Rightarrow N=\left(2-1\right)^{2018}+\left|1-1\right|^{2019}-\left(1-1\right)^{2020}=1\)
Sửa đề: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)
\(\Leftrightarrow x^2-2xy+y^2+y^2-2y+1+z^2-4z+4=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)
=>x=y=1 và z=2
\(A=\left(x-1\right)^{2018}+\left(y-1\right)^{2019}+\left(z-1\right)^{2020}\)
\(=\left(1-1\right)^{2018}+\left(1-1\right)^{2019}+\left(2-1\right)^{2020}\)
=1
Theo BĐT Cosi ta có: \(\hept{\begin{cases}\frac{x^4+y^4}{2}\ge\sqrt{x^4\cdot y^4}=x^2y^2\\\frac{y^4+z^4}{2}\ge\sqrt{y^4\cdot z^4}=y^2z^2\\\frac{z^4+x^4}{2}\ge\sqrt{z^4\cdot x^4}=x^2z^2\end{cases}\Rightarrow x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2}\)
chứng minh tương tự: \(x^2y^2+y^2z^2+z^2x^2\ge xy^2z+xyz^2+x^2yz\Leftrightarrow x^2y^2+y^2z^2+x^2z^2\ge xyz\left(x+y+z\right)\)
\(\Leftrightarrow x^2y^2+y^2z^2+x^2z^2\ge3xyz\)(do x+y+z=3)
Do đó: \(x^4+y^4+z^4\ge3xyz\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x^4=y^4;y^4=z^4;z^4=x^4\\x^2y^2=y^2z^2;y^2z^2=z^2x^2;z^2x^2=x^2y^2\end{cases}\Leftrightarrow x=y=z}\)(1)
mà x+y+z=3 (2)
Từ (1) và (2) => 3x=3 => x=1 => y=z=1
=> \(x^{2018}+y^{2019}+x^{2020}=1+1+1=3\)
\(2x^2+y^2+z^2-2xy-2x+1=0\)
\(\Rightarrow\left(x^2+y^2-2xy\right)+\left(x^2-2x+1\right)+z^2=0\)
\(\Rightarrow\left(x-y\right)^2+\left(x-1\right)^2+z^2=0\)
\(\Leftrightarrow x=y=1;=0\)
\(A=x^{2018}+y^{2019}+z^{2020}=1+1+0=2\)
2)
\(a+b+c=6\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=36\)
\(\Leftrightarrow12+2\left(ab+bc+ac\right)=36\Leftrightarrow ab+bc+ac=12\)
Kết hợp với \(a^2+b^2+c^2=12\Leftrightarrow a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
Kết hợp với \(a+b+c=6\Leftrightarrow a=b=c=2\)
\(P=\left(a-3\right)^{2019}+\left(b-3\right)^{2019}+\left(c-3\right)^{2019}=\left(-1\right)^{2019}+\left(-1\right)^{2019}+\left(-1\right)^{2019}=-3\)
Ta có: \(x^2+y^2+z^2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=6\)
<=> \(\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)+\left(z^2+\frac{1}{z^2}-2\right)=0\)
<=> \(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2+\left(z-\frac{1}{z}\right)^2=0\)
<=> \(\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\\z-\frac{1}{z}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{1}{x}\\y=\frac{1}{y}\\z=\frac{1}{z}\end{cases}}\)
<=> \(\hept{\begin{cases}x^2=1\\y^2=1\\z^2=1\end{cases}}\)
<=> x = y = z = \(\pm\)1
Với x = y = z = 1 => P = 12018 + 12019 + 12020 = 3
x = y = z = -1 => P = (-1)2018 + (-1)2019 + (-1)2020 = 1
Vậy ...
a) \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
b) \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37=100\)
c) \(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10=25\)
a) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
Ta có: \(\hept{\begin{cases}x^{2019}\le x^{2020}\\y^{2019}\le y^{2020}\end{cases}}\)
\(\Rightarrow x^{2019}+y^{2019}\le x^{2020}+y^{2020}\)
( em ko biết đúng hay sai làm theo cách hiểu của em thôi )