\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)

\(\left(4-8x\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)

\(2\left(x-4\right)-3x\left(x-4\right)=0\)

\(\left(2-3x\right)\left(x-4\right)=0\)

\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

Lời giải:
a. Mệnh đề sai, vì $x^2\geq 0>-1$ với mọi $x\in\mathbb{R}$ theo tính chất bình phương 1 sosos.

Mệnh đề phủ định: $\forall x\in\mathbb{R}, x^2\neq -1$

b. Mệnh đề đúng, vì $x^2+x+2=(x+0,5)^2+1,75>0$ với mọi $x\in\mathbb{R}$ nên $x^2+x+2\neq 0$ với mọi $x\in\mathbb{R}$

Mệnh đề phủ định: $\exists x\in\mathbb{R}| x^2+x+2=0$

\(=>x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1=\dfrac{13}{3}\)

\(=>x=\dfrac{13}{3}:\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{13}{3}:\dfrac{7}{6}=\dfrac{26}{7}\)

5/6x - 1 = 10/3

5/6x = 10/3 + 1 = 13/3

X = 13/3 : 5/6 = 26/5

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

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{10}{3}+1\)

\(\Rightarrow\dfrac{7}{6}x=\dfrac{13}{3}\)

\(\Rightarrow x=\dfrac{13}{3}:\dfrac{7}{6}\)

\(\Rightarrow x=\dfrac{26}{7}\)

a)x+x+2.x=76:2

=>2x+2x=38

=>2(x+x)=38

=>2x=19

=>x=19/2

b)2.x=x+14

=>2x-x=14

=>x=14

c)2.x-3/5=3/4

=>2x=3/4+3/5

=>2x=27/20

=>x=27/40

d)3/5-2.x=1/4

=>2x=3/5-1/4

=>2x=7/20

=>x=7/40

/2.x-3/=x-1 \(\Leftrightarrow\orbr{\begin{cases}2x-3=x-1\\2x-3=-(x-1)\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x-3-x=-1\\2x-3=-x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x(2-1)=2\\2x+x=4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\3x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}}\)

3 + x - 5 = 6x -4

3 - 5 + 4 = 6x -x 

2             = 5x 

x = 2 / 5 

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

\(3+x-5=6x-4\)

\(x-6x=-4+5-3\)

\(-5x=-2\)

\(x=\frac{2}{5}\)

3 + ( x - 5 ) = 2 ( 3x - 2 )

3 + ( x - 5 ) = 2 . 3x - 2. 2

3 + x + ( -5 ) = 6x - 4

3 + ( -5 ) + x = 6x - 4

-2 + x = 6x - 4

- 6x + x = 2 - 4

- 5 x = -2

x = -2 : ( -5 ) 

x = 0,4

\(\dfrac{7}{12}< \dfrac{x}{24}< \dfrac{2}{3}\) 

→ \(\dfrac{7}{12}=\dfrac{7x2}{12x2}=\dfrac{14}{24}\)

→ \(\dfrac{2}{3}=\dfrac{2x8}{3x8}=\dfrac{16}{24}\)
⇒ x\(=\)15

\(\Leftrightarrow14< x< 16\)

hay x=15

x=15

k nhé

please

<=> x³ - 4x² - x + 4 = 0

<=> (x² - 1)(x - 4) = 0

<=> (x - 1)(x + 1)(x - 4) = 0

<=> x = 1 hoặc x = -1 hoặc x = 4

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

\(\left(x-1\right)\left(x-4\right)\left(x+1\right)=0\)

\(x=\pm1;x=4\)

sắp tới lập nhóm nha :v ( quảng cáo lun ) 

\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)ĐKXĐ : \(x\ne\pm2\)

\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)

\(\Leftrightarrow x^2-4x+4+3x+6-x^2+11=0\)

\(\Leftrightarrow-x+21=0\)

\(\Leftrightarrow x=21\)( thỏa )

Vậy....