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:

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

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

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)

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

\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)

\(=-\left(2x-4\right)\left(x+8\right)\)

b) \(x^3+x^2y-15x-15y\)

\(=x^2\left(x+y\right)-15\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-15\right)\)

c) \(3\left(x+8\right)-x^2-8x\)

\(=3\left(x+8\right)-x\left(x+8\right)\)

\(=\left(x+8\right)\left(3-x\right)\)

d) \(x^3-3x^2+1-3x\)

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

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

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

\(=\left(x+1\right)\left(x^2-4x+1\right)\)

d) \(5x^2-5y^2-20x+20y\)

\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y-4\right)\)

Mấy cái số 2 đằng sau ngoặc là mũ đấy các bạn 

\(a,3(x-1)^2-3x(x-5)=1\)

\(\Leftrightarrow3x^2-6x+3-3x^2-15x=1\)

\(\Leftrightarrow\left[3x^2-3x^2\right]+3-\left[15x-6x\right]=1\)

\(\Leftrightarrow3-9x=1\)

\(\Leftrightarrow9x=2\Leftrightarrow x=\frac{2}{9}\)

\(c,(2x-5)(2x-5)-1=0\)

\(\Leftrightarrow4x^2-20x+25=1\)

\(\Leftrightarrow(2x-5)^2=1\)

\(\Leftrightarrow(2x-5)^2=1^2=(-1)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

\(3x^2-\left(3m-2\right)x-\left(3m+1\right)=0\left(1\right)\)\(\left(ĐK:a\ne0\right)\)

Theo phương trình ( 1 ) ta có:

\(\Delta=\left(3m-2\right)^2+4.3.\left(3m+1\right)\)

\(\Delta=9m^2-12m+4+36m+12\)

\(\Delta=9m^2+24m+16\)

\(\Delta=\left(3m\right)^2+2.3.4m+4^2=\left(3m+4\right)^2\)

Phương trình ( 1 ) có 2 nghiệm \(x_1;x_2\Leftrightarrow\Delta=\left(3m+4\right)^2>0\)

Mà \(\left(3m+4\right)^2\ge0\Rightarrow\left(3m+4\right)^2\ne0\)\(\Rightarrow3m\ne-4\Rightarrow m\ne-\frac{4}{3}\)

Ta có:  \(x_1+x_2=\frac{3m-2}{3}\left(2\right)\)

\(x_1-x_2=\frac{-3m-1}{3}\left(3\right)\)

\(3x_1-5x_2=6\left(2\right)\)

Từ ( 2 ) và ( 3 ) suy ra \(6x_2=\frac{3m-2}{3}-6\)\(\Rightarrow x_2=\frac{3m-2}{18}-1\)

Rồi làm tương tự với \(x_2\) tiếp tục thay \(x_1,x_2\)và phương trình ( 1 ) 

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

có \(\Delta=\left[-\left(3m-2\right)\right]^2-4.3.\left[-\left(3m+1\right)\right]\)

\(\Delta=9m^2-12m+4+36m+12\)

\(\Delta=9m^2+24m+16\)

\(\Delta=\left(3m+4\right)^2\ge0\forall m\)

vì theo đề bài để pt có 2 nghiệm nên thỏa mãn đk \(\forall m\)

ta có vi - ét \(\hept{\begin{cases}x_1+x_2=\frac{3m-2}{3}\left(1\right)\\x_1.x_2=-\frac{\left(3m+1\right)}{3}\left(2\right)\end{cases}}\)

theo bài ra \(3x_1-5x_2=6\)  \(\left(3\right)\)

từ \(\left(1\right),\left(3\right)\)  ta có hệ phương trình \(\hept{\begin{cases}x_1+x_2=\frac{3m-2}{3}\\3x_1-5x_2=6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x_1+x_2=m+\frac{2}{3}\\x_1-\frac{5}{3}x_2=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{8}{3}x_2=m+\frac{2}{3}-2\\x_1+x_2=m+\frac{2}{3}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x_2=\frac{3}{8}m-\frac{1}{2}\\x_1+\frac{3}{8}m-\frac{1}{2}=m+\frac{2}{3}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x_2=\frac{3}{8}m-\frac{1}{2}\\x_1=m-\frac{3}{8}m+\frac{2}{3}+\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x_2=\frac{3}{8}m-\frac{1}{2}\\x_1=\frac{5}{8}m+\frac{7}{6}\end{cases}}\)  \(\left(4\right)\)

thay (4) vào (2) ta được

\(\left(\frac{3}{8}m-\frac{1}{2}\right)\left(\frac{5}{8}m+\frac{7}{6}\right)=\frac{-3m-1}{3}\)

\(\Leftrightarrow\frac{15}{64}m+\frac{7}{16}-\frac{5}{16}m-\frac{7}{12}=-m-\frac{1}{3}\)

\(\Leftrightarrow\frac{-5}{64}m-\frac{7}{48}+m+\frac{1}{3}=0\)

\(\Leftrightarrow\frac{59}{64}m+\frac{3}{16}=0\)

\(\Leftrightarrow\frac{59}{64}m=\frac{-3}{16}\)

\(\Leftrightarrow m=\frac{-12}{59}\)  ( TM \(\forall m\))

vậy \(m=\frac{-12}{59}\)  là giá trị cần tìm 

\(1,\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

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

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)