a: ĐKXĐ: x<>2; x<>0

b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)

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

\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)

\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7

a) 2x+8≤ 0 

⇔2x≤-8

⇔x≤-4

b) 4x-7 ≥ 2x -5

⇔2x-12 ≥ 0

⇔2x≥12

⇔x≥6

c) (2x-8)(15-3x)>0

TH1: 2x-8>0 ⇒x>4

        15-3x>0⇒x<5 

TH2:  2x-8<0 ⇒x<4

        15-3x<0⇒x>5 (vô lí)

vậy 4<x<5

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

2x+1 – x = 3584 

x+1 = 3584 

x+14 

x = 4-1 

x = 3 

Chúc bạn học tốt 

Vì:  |2\(x\) - 1| = |1 - 2\(x\)|

Nên:      |2\(x\) - 1| + |1 - 2\(x\)| = 8

  ⇒        |2\(x\) - 1| + |2\(x\) - 1| = 8

                         2.|2\(x\) - 1| = 8

                            |2\(x\) - 1| = 8:2

                           |2\(x\) - 1| = 4

                         \(\left[{}\begin{matrix}2x-1=-4\\2x-1=4\end{matrix}\right.\)

                          \(\left[{}\begin{matrix}2x=-4+1\\2x=4+1\end{matrix}\right.\)

                            \(\left[{}\begin{matrix}2x=-3\\2x=5\end{matrix}\right.\)

                             \(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\){- \(\dfrac{3}{2}\); \(\dfrac{5}{2}\)}

giúp mình với

\(-4x^2+20-16x+4x^2=-3\)

\(\left(-4x^2+4x^2\right)-16x=-3-20\)

\(0-16x=-23\)

\(x=\frac{23}{16}\)

CHÚC BN HX TỐT :))

\(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)

\(\Leftrightarrow-4x^2+20x-16x+4x^2=-3\)

\(\Leftrightarrow4x=-3\Leftrightarrow x=-\frac{3}{4}\)

Bài 1L

a) \(\left(x-7\right)\left(x+3\right)< 0\)

TH1:

\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )

TH2:

\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )

Vậy \(-3< x< 7\)

Bài 2:

a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)

\(\Leftrightarrow5x+8-2x+15+21=2x-5\)

\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)

\(\Leftrightarrow x=-49\)

Vậy ...

Bài 1:

b) \(\left(2x+6\right)\left(x-5\right)\le0\)

TH1:

\(\hept{\begin{cases}2x+6\le0\\x-5\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le-3\\x\ge5\end{cases}}}\)( loại )

TH2:

\(\hept{\begin{cases}2x+6\ge0\\x-5\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\le5\end{cases}\Leftrightarrow}-3\le x\le5}\)

Vậy \(-3\le x\le5\)