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Ta có: \(\left|x+3\right|+\left|x-1\right|=\left|x+3\right|+\left|1-x\right|\ge\left|x+3+1-x\right|=4\)
\(\left|y-2\right|+\left|y+2\right|=\left|2-y\right|+\left|y+2\right|\ge\left|2-y+y+2\right|=4\)
\(\Rightarrow\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\le\dfrac{16}{4}=4\Rightarrow\left|x+3\right|+\left|x-1\right|\ge\dfrac{6}{\left|y-2\right|+\left|y+2\right|}\)
Dấu '=' xảy ra <=> (x+3)(1-x)\(\ge0\) và (2-y)(y+2)\(\ge0\)
Vì x,y \(\in Z\Rightarrow\left\{{}\begin{matrix}x\in\left\{-3;-2;-2;0;1\right\}\\y\in\left\{-2;-1;0;1;2\right\}\end{matrix}\right.\)
a: \(x^3-2y^2=2^3-2\cdot\left(-2\right)^2=8-2\cdot4=0\)
=>\(C=x\left(x^2-y\right)\left(x^3-2y^2\right)\left(x^4-3y^3\right)\left(x^5-4y^4\right)=0\)
b: x+y+1=0
=>x+y=-1
\(D=x^2\left(x+y\right)-y^2\left(x+y\right)+\left(x^2-y^2\right)+2\left(x+y\right)+3\)
\(=x^2\cdot\left(-1\right)-y^2\left(-1\right)+\left(x^2-y^2\right)+2\cdot\left(-1\right)+3\)
\(=-x^2+y^2+x^2-y^2-2+3\)
=1
Ta có \(\left|y-1\right|+\left|y-2\right|+\left|y-3\right|+1=\left|y-1\right|+\left|y-2\right|+\left|3-y\right|+1\ge2+\left|y-2\right|+1=3+\left|y-2\right|\ge3\)
\(\dfrac{6}{\left(x-1\right)^2+2}\le\dfrac{6}{0+2}=3\)
\(\Leftrightarrow VT\le3\le VP\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\left(y-1\right)\left(3-y\right)\ge0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy PT có nghiệm \(\left(x;y\right)=\left(1;2\right)\)
Tìm x,y biết
\(\dfrac{6}{\left(x-1\right)^2+2}=\left|y-1\right|+\left|y-2\right|+\left|y-3\right|+1\)
\(1)\)
\(VT=\left(\left|x-6\right|+\left|2022-x\right|\right)+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(\ge\left|x-6+2022-x\right|+\left|0\right|+\left|0\right|+\left|0\right|=2016\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-6\right)\left(2022-x\right)\ge0\left(1\right)\\x-10=y-2014=z-2015=0\left(2\right)\end{cases}}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=10\\y=2014\\z=2015\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}x-6\ge0\\2022-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge6\\x\le2022\end{cases}\Leftrightarrow}6\le x\le2022}\) ( nhận )
TH2 : \(\hept{\begin{cases}x-6\le0\\2022-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le6\\x\ge2022\end{cases}}}\) ( loại )
Vậy \(x=10\)\(;\)\(y=2014\) và \(z=2015\)
\(2)\)
\(VT=\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=\left|-4\right|=4\)
\(VP=\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\)
\(\Rightarrow\)\(VT\ge VP\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-5\right)\left(1-x\right)\ge0\left(1\right)\\\left|y+1\right|=0\left(2\right)\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}x-5\ge0\\1-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le1\end{cases}}}\) ( loại )
TH2 : \(\hept{\begin{cases}x-5\le0\\1-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le5\\x\ge1\end{cases}\Leftrightarrow}1\le x\le5}\) ( nhận )
\(\left(2\right)\)\(\Leftrightarrow\)\(y=-1\)
Vậy \(1\le x\le5\) và \(y=-1\)
Ta có: \(\left(x+y-2\right)^2+7\ge7\Rightarrow\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\ge7\)
\(\Rightarrow\left|y-1\right|+\left|y-3\right|\le2\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|y-1\right|=0\\\left|y-3\right|=2\end{matrix}\right.\\\left\{{}\begin{matrix}\left|y-1\right|=2\\\left|y-3\right|=0\end{matrix}\right.\\\left|y-1\right|=\left|y-3\right|=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1\\y=3\\y=2\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)
a) \(\Leftrightarrow\left|x-3\right|=0;\left|y-2x\right|=0;\left|2z-x+y\right|=0\)
\(\Leftrightarrow x=3;y=2x;2z=-y+x\)
Ta có : y = 2x => y = 2 . 3 = 6
và 2z = -y + x => 2z = -6 + 3 = -3 => z = \(-\frac{3}{2}\)
b) \(\Leftrightarrow\left|x-y\right|+\left|2y+x-\frac{1}{2}\right|+\left|x+y+z\right|=0\) (vĩ mỗi số hạng trong tổng đều lớn hơn hoặc bằng 0)
\(\Leftrightarrow\left|x-y\right|=0;\left|2y+x-\frac{1}{2}\right|=0;\left|x+y+z\right|=0\)
\(\Leftrightarrow x=y;2y+x=\frac{1}{2};x+y=-z\)
Vì x = y nên \(2y+x=3y=\frac{1}{2}\Rightarrow x=y=\frac{1}{2}:3=\frac{1}{6}\)
và \(-z=x+y=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\Rightarrow z=-\frac{1}{3}\)
Lời giải:
Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:
$|x-1|+|x-4|=|x-1|+|4-x|\geq |x-1+4-x|=3$
$|x-2|+|y-3|\geq 0$
$\Rightarrow |x-1|+|x-2|+|y-3|+|x-4|\geq 3$
Dấu "=" xảy ra khi:
\(\left\{\begin{matrix}
(x-1)(4-x)\geq 0\\
x-2=0\\
y-3=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x=2\\
y=3\end{matrix}\right.\)