a: =>2x>-6

hay x>-3

e: =>(5-x)/x<0

=>0<x<5

h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)

\(\Leftrightarrow x+3< 0\)

hay x<-3

g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)

Đặt F(\(x\)) = (\(x\) - 1)(\(x\)+3)(\(x\) - 4)>0

Lập bảng xét dấu:

Theo bảng trên ta có Nghiệm của bất phương trình là:

\(\left[{}\begin{matrix}x\in\left\{-2;-1;0\right\}\\x\in\left\{x\in Z/x>4\right\}\end{matrix}\right.\)

\(x:\dfrac{3}{4}\) = \(\dfrac{9}{14}-\dfrac{1}{7}\)

\(x:\dfrac{3}{4}\) = \(\dfrac{1}{2}\)

\(x=\dfrac{1}{2}\) x \(\dfrac{3}{4}\)

\(x=\dfrac{3}{8}\)

\(x\)x \(\dfrac{1}{2}-\dfrac{1}{4}\)=\(\dfrac{2}{3}\)

\(x\) x \(\dfrac{1}{4}\)     =\(\dfrac{2}{3}\)

\(x\)            = \(\dfrac{2}{3}:\dfrac{1}{4}\)

\(x\)             \(\dfrac{8}{3}\)

x=17 nhé

mk kô bt đánh ps đâuu à

a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)

\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)

\(\Leftrightarrow x-4=25\)

\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)

b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=18.4\)

\(\Leftrightarrow x\left(x+1\right)=72\)

vì \(72=8.9=\left(-8\right).\left(-9\right)\)

\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)

c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)

\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)

\(\Leftrightarrow2x+3-2x-8⋮x+4\)

\(\Leftrightarrow-5⋮x+4\)

\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)

Câu 2:

\(A\left(x\right)=x^2+3x+1\)

\(B\left(x\right)=2x^2-2x-3\)

a) Tính A(x) là sao em?

b) \(A\left(x\right)+B\left(x\right)=\left(x^2+3x+1\right)+\left(2x^2-2x-3\right)\)

\(=x^2+3x+1+2x^2-2x-3\)

\(=\left(x^2+2x^2\right)+\left(3x-2x\right)+\left(1-3\right)\)

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

Câu 1:

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

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

\(=x+2\)

Bậc của M(x) là 1