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: \(\left|x+3\right|+\left|x-1\right|=\left|x+3\right|+\left|1-x\right|\ge\left|x+3+1-x\right|=4\)

\(\left|y-2\right|+\left|y+2\right|=\left|2-y\right|+\left|y+2\right|\ge\left|2-y+y+2\right|=4\)

\(\Rightarrow\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\le\dfrac{16}{4}=4\Rightarrow\left|x+3\right|+\left|x-1\right|\ge\dfrac{6}{\left|y-2\right|+\left|y+2\right|}\)

Dấu '=' xảy ra <=> (x+3)(1-x)\(\ge0\) và (2-y)(y+2)\(\ge0\)

Vì x,y \(\in Z\Rightarrow\left\{{}\begin{matrix}x\in\left\{-3;-2;-2;0;1\right\}\\y\in\left\{-2;-1;0;1;2\right\}\end{matrix}\right.\)

C

B

\(y'=3\left(x^2-\dfrac{2}{x}\right)^2.\left(x^2-\dfrac{2}{x}\right)'=3\left(x^2-\dfrac{2}{x}\right)^2\left(2x+\dfrac{2}{x^2}\right)\)

\(=6\left(x+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{2}{x}\right)^2\)

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) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{6}{\left(x-2\right)^2+3}=2\\\left|y-1\right|+2=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left|y-1\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

Ta có: \(\left(x+y-2\right)^2+7\ge7\Rightarrow\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\ge7\)

\(\Rightarrow\left|y-1\right|+\left|y-3\right|\le2\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|y-1\right|=0\\\left|y-3\right|=2\end{matrix}\right.\\\left\{{}\begin{matrix}\left|y-1\right|=2\\\left|y-3\right|=0\end{matrix}\right.\\\left|y-1\right|=\left|y-3\right|=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1\\y=3\\y=2\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)

de bai

tìm x,y

a) Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có :

\(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\)

\(\ge\left|x+2+1-x\right|=3\) (1)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow-2\le x\le1\)

+ \(\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow3-\left(y+2\right)^2\le3\) (2)

Dấu "=" xảy ra \(\Leftrightarrow\left(y+2\right)^2=0\Leftrightarrow y=-2\)

Từ (1) và (2) suy ra \(\left|x+2\right|+\left|x+1\right|=3-\left(y+2\right)^2=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le1\\y=-2\end{matrix}\right.\)

b) \(\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\) (3)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow1\le x\le5\)

+ \(\left|y+1\right|\ge0\forall y\) \(\Rightarrow\left|y+1\right|+3\ge3\)

\(\Rightarrow\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\) (4)

Dấu "=" xảy ra \(\Leftrightarrow\left|y+1\right|=0\Leftrightarrow y=-1\)

Từ (3) và (4) suy ra \(\left|x-5\right|+\left|1-x\right|=\frac{12}{\left|y+1\right|+3}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)

Câu c,d lm tương tự