Khi x > 9 thì 7 - x < 0 do đó |7-x| = x - 7 

Khi đó M=4+2x+l7-xl = 4 + 2x + x- 7 = 3x -3

Vậy M = 3x - 3

\(\dfrac{3\left(x-1\right)}{4}+1< \dfrac{x+2}{3}\)

\(\Leftrightarrow\dfrac{9\left(x-1\right)}{12}+\dfrac{12}{12}< \dfrac{4\left(x+2\right)}{12}\)

\(\Rightarrow9\left(x-1\right)+12< 4\left(x+2\right)\)

\(\Leftrightarrow9x-9+12< 4x+8\)

\(\Leftrightarrow9x+3< 4x+8\)

\(\Leftrightarrow9x-4x< 8-3\)

\(\Leftrightarrow5x< 5\)

\(\Leftrightarrow x< 1\)

x+600= \(\dfrac{180x}{100}\)- 360
x- \(\dfrac{180x}{100}\)= -600- 360
\(\dfrac{-80x}{100}\)= -960
-80x= -960.100
x= 1200

Câu e,d à bạn

bài1 th ạ

a) \(5x-20=0\)

\(\Leftrightarrow5x=20\)

\(\Leftrightarrow x=4\)

b) \(\left(3x+1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=5\end{matrix}\right.\)

c) \(\dfrac{2\left(x^2-2\right)}{x^2-4}-\dfrac{x-1}{x+2}=\dfrac{x+1}{x-1}\left(ĐK:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{2x^2-4}{x^2-4}-\dfrac{\left(x-1\right)\left(x-2\right)}{x^2-4}=\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}\)

\(\Rightarrow2x^2-4-\left(x^2-2x-x+2\right)=x^2+2x+x+2\)

\(\Leftrightarrow2x^2-4-x^2+3x-2-x^2-3x-2=0\)

\(\Leftrightarrow\left(2x^2-x^2-x^2\right)+\left(3x-3x\right)-8=0\)

\(\Leftrightarrow-8=0\) (Vô lý)

d) \(2x+3< x-5\)

\(\Leftrightarrow2x-x< -5-3\)

\(\Leftrightarrow x< -8\)

e) \(2\left(x+1\right)-5x\ge x+7\)

\(\Leftrightarrow2x+2-5x\ge x+7\)

\(\Leftrightarrow2x-5x-x\ge7-2\)

\(\Leftrightarrow-4x\ge5\)

\(\Leftrightarrow x\le-\dfrac{5}{4}\)

thoyy,tui xin chịu ạ

=1/4+1/5-1/5+...-1/39+1/40

=1/4+1/19

=.....(tự tính nka)

Ta có:

\(\left|x-3\right|=x-3\Leftrightarrow x-3\ge0\Leftrightarrow x\ge3\)

\(\left|x-3\right|=-x+3\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

Với \(x\ge3\Rightarrow x-3=9-2x\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=4\) (Thoả mãn)

Với \(x< 3\Rightarrow-x+3=9-2x\)

\(\Leftrightarrow x=6\) (Loại)