NN
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28 tháng 9 2021
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
TN
3
MV
7 tháng 1 2018
\(\left|x-5\right|+\left|x-1\right|=\left|5-x\right|+\left|x-1\right|\ge\left|5-x+x-1\right|=\left|4\right|=4\\ \left|y+5\right|\ge0\\ \Leftrightarrow\left|y+5\right|+3\ge3\\ \Leftrightarrow\dfrac{12}{\left|y+5\right|+3}\le\dfrac{12}{3}=4\\ VT\ge4;VP\le4\\ \Rightarrow\text{Dấu "=" phải xảy ra }\Leftrightarrow\left\{{}\begin{matrix}\left(5-x\right)\left(x-1\right)\ge0\\y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-5\end{matrix}\right.\)
NN
7 tháng 1 2018
Ta có: \(\left\{{}\begin{matrix}\left|x-5\right|+\left|x-1\right|=\left|5-x\right|+\left|x-1\right|\ge\left|5-x+x-1\right|\ge\left|4\right|=4\\\dfrac{12}{\left|y+5\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right..\)
\(\Rightarrow\left|x-5\right|+x-1=\dfrac{12}{\left|y+5\right|+3}=4.\)
\(\Rightarrow\left\{{}\begin{matrix}1< x< 5\\y=-5\end{matrix}\right..\)
Vậy..........
23 tháng 3 2021
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
1 tháng 12 2021
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 \(\left(x;y\right)=\left(1;2\right)\)
DH
Tìm x,y biết
\(\dfrac{6}{\left(x-1\right)^2+2}=\left|y-1\right|+\left|y-2\right|+\left|y-3\right|+1\)
0
NN
1
20 tháng 2 2021
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 \(\in Z\Rightarrow\left\{{}\begin{matrix}x\in\left\{-3;-2;-2;0;1\right\}\\y\in\left\{-2;-1;0;1;2\right\}\end{matrix}\right.\)
Ta có: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Dấu '=' xảy ra <=> \(ab\ge0\)
Lại có: \(\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\Rightarrow\left|x-5\right|+\left|1-x\right|\ge4\ge\dfrac{12}{\left|y+1\right|+3}\)
Đẳng thức xảy ra <=> \(\left(x-5\right)\left(1-x\right)\ge0;y+1=0\Rightarrow y=-1\)
\(x\in Z\Rightarrow x\in\left\{5;4;3;2;1\right\}\)
Vậy ta có cặp số nguyên (x;y)=(5;-1),(4;-1),(3;-1),(2;-1),(1;-1)