\(VT=\left|3x+1\right|+\left|3x-5\right|=\left|3x+1\right|+\left|5-3x\right|\ge\left|3x+1+5-3x\right|=6\)

\(VP=\frac{12}{\left(y+3\right)^2+2}\le\frac{12}{2}=6\)

Như vậy \(VT\ge6;VP\le6\)

Mà \(VT=VP\Leftrightarrow VT=VP=6\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\frac{1}{3}\le x\le\frac{5}{3}\\y=-3\end{cases}}\)

Ta có: \(\left|3x+1\right|+\left|3x-5\right|=\left|3x+1\right|+\left|5-3x\right|\ge\left|3x+1+5-3x\right|=6\)(1)

\(\frac{12}{\left(y+3\right)^2+2}\le\frac{12}{2}=6\)(2)

\(\left(1\right);\left(2\right)\Rightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{1}{3}\le x\le\frac{5}{3}\\y=-3\end{cases}}\)

Thanks bn nha !! Nka

Lời giải:

Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:

$|3x+1|+|3x-5|=|3x+1|+|5-3x|\geq |3x+1+5-3x|=6$

$(y+3)^2+2\geq 2, \forall y\Rightarrow \frac{12}{(y+3)^2+2}\leq \frac{12}{2}=6$

Vậy:

$|3x+1|+|3x-5|\geq 6\geq \frac{12}{(y+3)^2+2}$
Dấu "=" xảy ra (3x+1)(5-3x)\geq 0$ và $y+3=0$

$\Leftrightarrow \frac{-1}{3}\leq x\leq \frac{5}{3}$ và $y=-3$

hic

\(1)\)

\(VT=\left(\left|x-6\right|+\left|2022-x\right|\right)+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(\ge\left|x-6+2022-x\right|+\left|0\right|+\left|0\right|+\left|0\right|=2016\)

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 \(y=-1\)

bài này sẽ giải nếu x,y là số nguyên

ĐKXĐ: x≠2

A=\(\dfrac{3\left(x++y\right)\left(x-2\right)+1}{x-2}\)

A=\(\dfrac{3\left(x+y\right)\left(x-2\right)}{x-2}+\dfrac{1}{x-2}\)

A=3(x+y)+\(\dfrac{1}{x-2}\)

Vì x;y; A là số nguyên nên \(\dfrac{1}{x-2}\) cũng là số nguyên

hay x-2⋮1

hay x-2ϵƯ(1)=(-1;1)

suy ra x=1;3

tự tìm y