a: 3(x+7)-2x+5>0

=>3x+21-2x+5>0

=>x+26>0

=>x>-26

Sửa đề: \(\dfrac{x+2}{18}-\dfrac{x+3}{8}< \dfrac{x-1}{9}-\dfrac{x-4}{24}\)

=>\(\dfrac{4\left(x+2\right)}{72}-\dfrac{9\left(x+3\right)}{72}< \dfrac{8\left(x-1\right)}{72}< \dfrac{3\left(x-4\right)}{72}\)

=>\(4\left(x+2\right)-9\left(x+3\right)< 8\left(x-1\right)-3\left(x-4\right)\)

=>\(4x+8-9x-27< 8x-8-3x+12\)

=>-5x-19<5x+4

=>-10x<23

=>\(x>-\dfrac{23}{10}\)

b: \(3x+2+\left|x+5\right|=0\left(1\right)\)

TH1: x>=-5

(1) trở thành: 3x+2+x+5=0

=>4x+7=0

=>\(x=-\dfrac{7}{4}\left(nhận\right)\)

TH2: x<-5

=>x+5<0

=>|x+5|=-x-5

Phương trình (1) sẽ trở thành:

\(3x+2-x-5=0\)

=>2x-3=0

=>2x=3

=>\(x=\dfrac{3}{2}\)

\(\dfrac{x+5}{x-5}=\dfrac{5}{x^2-5x}+\dfrac{1}{x}\)

\(\Leftrightarrow\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x-5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)

Ta có : \(\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)

\(\Leftrightarrow\dfrac{x\left(x+5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}+\dfrac{x-5}{x\left(x-5\right)}\)

`=> x (x+5) = 5 +x-5`

`<=> x^2 +5x - 5-x+5=0`

`<=> x^2 +4x =0`

`<=> x(x+4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)

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

Vậy phương trình có nghiệm `x=-4`

x(x²+6x+9) - 3x= x³+6x²+12x+8+1

\(\Leftrightarrow\)x³+6x²+9x-3x=x³+6x²+12x+9

\(\Leftrightarrow\)6x=12x+9

\(\Leftrightarrow\)6x=-9

\(\Leftrightarrow\)x=-3/2

Vậy phương trình có 1 nghiệm duy nhất x=-3/2

x(x + 3)^2  - 3x = (x + 2)^3 + 1

<=> x(x^2 + 6x + 9) = x^3 + 6x^2 + 12x + 8 + 1

<=> x^3 + 6x^2 + 9x = x^3 + 6x^2 + 12x + 9

<=> 3x + 9 = 0

<=> 3x = -9

<=> x = -3

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

\(\Leftrightarrow x\left(x^2+6x+9\right)=x^3+6x^2+12x+8+1\)

\(\Leftrightarrow x^3+6x^2+9x=x^3+6x^2+12x+9\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\)

\(x+x^2=0\)

\(\Leftrightarrow x\left(1+x\right)=0\)

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

Vậy: Phương trình có tập nghiệm \(S=\left\{0;-1\right\}\)

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

`5/(-x^2+5x-6)+(x+3)/(2-x)=0`

Đk:`x ne 2,x ne 3`

`pt<=>-5/(x^2-5x+6)-(x+3)/(x-2)=0`

`<=>-5-(x+3)(x-3)=0`

`<=>(x+3)(x-3)=-5`

`<=>x^2-9=-5`

`<=>x^2-4=0`

`<=>(x-2)(x+2)=0`

`x ne 2=>x-2 ne 0`

`<=>x+2=0`

`<=>x=-2`

Vậy `S={-2}`

Tham khảo thử đúng không nha mn

     \(x^2+x-y^2=0\)

⇔ \(\left(x^2-y^2\right)+x=0\)

⇔ \(\left(x-y\right)\left(x+y\right)+x=0\)

⇒ \(x-y=0\) hoặc \(x+y=0\) hoặc \(x=0\)

⇒ \(x=y=0\)

\(\Leftrightarrow x^2+x=y^2\)

\(\Leftrightarrow4x^2+4x=4y^2\)

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

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

\(\Leftrightarrow\left(2x-2y+1\right)\left(2x+2y+1\right)=1\)

Vậy \(\left(x;y\right)=\left(0;0\right);\left(-1;0\right)\)

\(x\left(x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x=0\) hay \(x+3=0\) hay \(x^2+1=0\) (pt vô nghiệm vì \(x^2+1\ge1\))

\(\Leftrightarrow x=0\) hay \(x=-3\)

-Vậy \(S= \left\{0;-3\right\}\)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)