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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
\(\dfrac{6}{\left(x-1\right)^2+2}=\left|y-1\right|+\left|y-2\right|+\left|y-3\right|+1\)
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 \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
Ta có: \(\left(x+y-2\right)^2+7\ge7\Rightarrow\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\ge7\)
\(\Rightarrow\left|y-1\right|+\left|y-3\right|\le2\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|y-1\right|=0\\\left|y-3\right|=2\end{matrix}\right.\\\left\{{}\begin{matrix}\left|y-1\right|=2\\\left|y-3\right|=0\end{matrix}\right.\\\left|y-1\right|=\left|y-3\right|=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1\\y=3\\y=2\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
Do \(\left(x-2\right)^2\ge0;\forall x\Rightarrow\dfrac{6}{\left(x-2\right)^2+3}\le\dfrac{6}{0+3}=2\) (1)
\(\left|y-1\right|\ge0;\forall y\Rightarrow\left|y-1\right|+2\ge2\) (2)
Từ (1); (2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{6}{\left(x-2\right)^2+3}=2\\\left|y-1\right|+2=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left|y-1\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=1.6=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
c: =>x-y=0 và y+9/25=0
=>x=y=-9/25
d: =>-1/3<x-3/5<1/3
=>4/15<x<14/15
e: =>|x+5,5|>5,5
=>x+5,5>5,5 hoặc x+5,5<-5,5
=>x>0 hoặc x<-11
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 \(\in Z\Rightarrow\left\{{}\begin{matrix}x\in\left\{-3;-2;-2;0;1\right\}\\y\in\left\{-2;-1;0;1;2\right\}\end{matrix}\right.\)