a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

a, |x^2 - 3x| = 0

=> x^2 - 3x = 0

=> x(x - 3) = 0

=> x = 0 hoặc x - 3 = 0 

=> x = 0 hoặc x = 3

vậy_

\(\left|a^2-3a\right|=0\)

\(\Rightarrow a^2-3a=0\)

\(\Rightarrow a\left(a-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}a=0\\a=3\end{cases}}\)

Trả lời

Vì lx²-3xl=0

Nên x²-3x=0

=>x(x-3)=0

x=0.           x-3=0

                    x=3

Học tốt

Áp dụng Côsi

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2b\)

Tương tự: \(\frac{bc}{a}+\frac{ca}{b}\ge2c;\frac{ca}{b}+\frac{ab}{c}\ge2a\)

\(\Rightarrow2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)=2\)

\(\Rightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge1\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)

Vậy GTNN của A là 1

x=46 nha

ghi rõ giupd e vs

a)\(\frac{1}{3}+\frac{1}{2}:x=-4\)

\(\Leftrightarrow\frac{1}{2}:x=-\frac{13}{3}\)

\(\Leftrightarrow x=-\frac{3}{26}\)

b)\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4}:x=-\frac{7}{20}\)

\(\Leftrightarrow x=-\frac{5}{7}\)

#H

a) \(\left(x+1\right)\left(x-2\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-1< x< 2\\x\in\varnothing\end{matrix}\right.\) vậy \(-1< x< 2\)

b) \(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{-2}{3}\end{matrix}\right.\) vậy \(x>2\) hoặc \(x< \dfrac{-2}{3}\)

a) (-3).(x+2)<0

=>x+2>0

=>x> -2

b)(x-1).(x+\(\dfrac{1}{3}\))>0

<=>(x-1) và \(\left(x+\dfrac{1}{3}\right)\) cùng dấu

TH1: x-1 <0

và x+\(\dfrac{1}{3}\)<0

\(\left\{{}\begin{matrix}x< 1\\x< \dfrac{-1}{3}\end{matrix}\right.\) =>x<\(\dfrac{-1}{3}\)

TH2:x-1>0

và x+\(\dfrac{1}{3}\)>0

\(\left\{{}\begin{matrix}x>1\\x>\dfrac{-1}{3}\end{matrix}\right.\)=>x>1

a)(-3).(x+2) <0

=> x+2> 0

=> x>2

b)(x-1).(x+\(\dfrac{1}{3}\)) >0

=> x-1>0 hay x+\(\dfrac{1}{3}\) >0

=> x>1hay x>-\(\dfrac{1}{3}\)