5x² - 15x = 0

5x(x-3)=0

suy ra 2 trường hợp

x=0

x-3=0=>x=3

5x²-15x=0
5x(x-3) =0
TH1: 5x=0            TH2: x-3=0
       =>x=0                   => x=3
    Vậy x thuộc {0;3}

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=5\end{matrix}\right.\)

\(x^3-5x^2-x+5=x^2\left(x-5\right)-\left(x-5\right)=\left(x-5\right)\left(x^2-1\right)=\left(x-5\right)\left(x-1\right)\left(x+1\right)\)

x3+3x2-10x=0

=>x(3+3.2-10)=0

=>x=0

x3-5x2-14x=0

=>x(3-5.2-14)=0

=>x=0

x3+5x2-24x=0

=>x(3+5.2-24)=0

=>x=0

Câu a)

\(x^3+3x^2-10=0\Rightarrow x\left(x^2+3x-10\right)=0\Rightarrow x\left(x^2-2x+5x-10\right)=0\Rightarrow x\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\Rightarrow x\left(x+5\right)\left(x-2\right)=0\)

\(\Rightarrow x=0;x=5;x=2\)

Câu b:

\(x^3-5x^2-14x=0\Rightarrow x\left(x^2-5x-14\right)=0\Rightarrow x\left(x^2+2x-7x-14\right)=0\Rightarrow x\left(x\left(x+2\right)-7\left(x+2\right)\right)=0\Rightarrow x\left(x-7\right)\left(x+2\right)=0\)

\(\Rightarrow x=0;x=7;x=-2\)

Sửa đề: \(5x^2+5y^2+8xy-2x+2y+2=0\)

=>\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

=>\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

=>\(\left\{{}\begin{matrix}2x+2y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(M=\left(x-y\right)^{2023}-\left(x-2\right)^{2024}+\left(y+1\right)^{2023}\)

\(=\left(1+1\right)^{2023}-\left(1-2\right)^{2024}+\left(-1+1\right)^{2023}\)

\(=2^{2023}-1\)

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

Đề là như này ạ ? nếu là vậy bạn xem có đúng đề không ạ .

\(5x^2-15x=0\Leftrightarrow5x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ 3\left(x+5\right)-2x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)

1) \(5x^2-15x=0\)

\(\Rightarrow5x\left(x-3\right)=0\)

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

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

\(\Rightarrow\left(x+5\right)\left(3-2x\right)=0\)

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

\(5x^2-15x=0\)

\(\Rightarrow x.\left(5x-15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\5x=15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy x ∈ {0 ; 3}

\(3.\left(x+5\right)-2x.\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right).\left(3-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+5=0\\3-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy x ∈ \(\left\{-5;\dfrac{3}{2}\right\}\)

c) x ∈ Z và  x 2 - 5 x 2 - 24 < 0

Ta có:  x 2 - 5 > 0 ; x 2 - 24 < 0 ⇒ x 2 > 5 ; x 2 < 24 Nên  x 2 ∈ 9 ; 16

x 2 = 9 ⇒ x = ± 3 ; x = 16 ⇒ x = ± 4

Vậy  x ∈ - 3 ; 3 ; - 4 ; 4

\(\Leftrightarrow\left(x-3\right)\left(2x^2+2\right)+5x\left(3-x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\)

=>(x-3)(x-2)(2x-1)=0

=>x=3 hoặc x=2 hoặc x=1/2

\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2+2\right)-5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left[\left(2x^2-4x\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\)

\(\Leftrightarrow2x^3+2x-6x^2-6+15x-5x^2=0\)

\(\Leftrightarrow2x^3-11x^2+17x-6=0\)

\(\Leftrightarrow2x^3-4x^2-7x^2+14x+3x-6=0\)

\(\Leftrightarrow2x^2\left(x-2\right)-7x\left(x-2\right)+3\left(x-2\right)=0\)

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

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

\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x-3\right)=0\)

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

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

Vậy: \(S=\left\{2;\dfrac{1}{2};3\right\}\)

\(y+5x2-\left(32+16x3:6-15\right)=0\)

\(y+10-\left(32+8-15\right)=0\)

\(y+10-\left(40-15\right)=0\)

\(y+10-25=0\)

\(y=25-10\)

\(y=15\)