D.\(x^2+5x+9< 0\)

\(x^2+5x+9=\left(x^2+2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right)-\left(\dfrac{5}{2}\right)^2+9=\left(x+\dfrac{5}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)

Mà \(x^2+5x+9< 0\)

--> pt vô nghiệm

Ta có: \(x^3-5x^2+6x=0\)

\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)

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

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

Vậy: S={0;2;3}

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

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

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

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

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

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

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

\(S=\left\{0,2,3\right\}\)

x^3 -5x^2 + 6x=0

\(< =>x\left(x^2-5x+6\right)=0\)

\(< =>x=0;x^2-5x+6=0\)

+/ \(x^2-5x+6=0\)

\(< =>x^2-2x-3x+6=0\)

\(< =>\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(< =>x\left(x-2\right)-3\left(x-2\right)=0\)

\(< =>\left(x-2\right)\left(x-3\right)=0\)

=> \(x-2=0;x-3=0\)

=> \(x=2;3\)

Vayayj tập nghiệm của......

Link : Hoc24

Ta có:

     (2 - 3x)(x + 8) = (3x - 2)(3 - 5x)

⇔ (2 - 3x)(x + 8) - (3x - 2)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8) + (2 - 3x)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8 + 3 - 5x) = 0

⇔ (2 - 3x)(11 - 4x) = 0

⇔ 2 - 3x = 0 hay 11 - 4x = 0

⇔ 2 = 3x hay 11 = 4x

⇔ x = \(\dfrac{2}{3}\) hay x = \(\dfrac{11}{4}\)

Vậy tập nghiệm của pt S = \(\left\{\dfrac{2}{3};\dfrac{11}{4}\right\}\)

khum bt mới lớp 5 hoi à(k10)

<=> (2-3x ) (x+8) + (2-3x ) (3-5x)=0
<=> (2-3x ) ( x+8 +  3-5x ) =0 
<=> (2-3x ) ( 11 - 4x ) = 0
 => 2-3x  =0 hoặc 11-4x =0  
       3x = 2            4x =11
         x = 2/3         x    = 11/4

a: =>-x+2x=3-7

=>x=-4

b: =>6x+2+2x-5=0

=>8x-3=0

hay x=3/8

c: =>5x+2x-2-4x-7=0

=>3x-9=0

hay x=3

d: =>10x²-10x²-15x=15

=>-15x=15

hay x=-1

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

sửa đề : 

\(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\Leftrightarrow x=1\)

a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)

b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)

d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)

a) Ta có: 4x-20=0

$\iff 4x=20$

hay x=5

Vậy: S={5}

b) Ta có: $2x+x+12=0$

$\iff 3x+12=0$

$\iff 3x=-12$

hay x=-4