Ta có: \(\left|y+3\right|\ge0\forall y\)

\(\left|2x+y\right|\ge0\forall x,y\)

Do đó: \(\left|y+3\right|+\left|2x+y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}y+3=0\\2x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-3\end{matrix}\right.\)

a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)

b: Ta có: |2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>x=-2

Thay x=-2 vào A, ta được:

\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)

c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)

=>x+3=24x+12

=>24x+12=x+3

=>23x=-9

hay x=-9/23

d: Để A<0 thì x+3<0

hay x<-3

Ta có: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{7}{6}\end{matrix}\right.\)

50 x 125 x 2 x 8 x 805

(50 . 2).(125. 8). 805

100. 1000 . 805

100000. 805

80500000

chúc bạn học tốt!!!

50 x 125 x 2 x 8 x 805

= ( 50 x 2 ) x ( 125 x 8 ) x 805

= 100 x 1000 x 805

= 100 000 x 805

= 80 500 000

_HT_

50x125x2x8x805=50x2x125x8x805=100x1000x805=80500000          ok em nhé

Ta có: \(\dfrac{x+1}{2018}+\dfrac{x+1}{2019}+\dfrac{x+1}{2020}+\dfrac{x+1}{2021}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

17 + (x - 7) + 2(2x + 1) = 17

<=> x - 7 + 2(2x + 1) = 17 - 17

<=> x - 7 + 2(2x + 1) = 0

<=> x + 2(2x + 1) = 0 + 7

<=> x + 2(2x + 1) = 7

<=> x + 4x + 1 = 7

<=> 5x + 2 = 7

<=> 5x = 7 - 2

<=> 5x = 5

=> x = 1

\(17+\left(x-7\right)+2\left(2x+1\right)=17\)

\(\Rightarrow x-7+4x+2=0\)

\(\Rightarrow5\left(x-1\right)=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(17+\left(x-7\right)+2\left(2x+1\right)=17\)

\(\Leftrightarrow17+x-7+4x+2=17\)

\(\Leftrightarrow\left(17-7\right)+\left(x+2x\right)+2=17\)

\(\Leftrightarrow10+3x+2=17\)

\(\Leftrightarrow\left(10+2\right)+3x=17\)

\(\Leftrightarrow12+3x=17\)

\(\Leftrightarrow3x=17-12\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\frac{5}{3}\)