a) Ta có: \(x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

b) Ta có: \(x^2+3x-18=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

X^3-11x^2+30x=0

<=>x(x²-11x+30)=0

<=>x(x²-5x-6x+30)=0

<=>x[x(x-5)-6(x-5)]=0

<=>x(x-5)(x-6)=0

<=>x=0 hoặc x=5 hoặc x=6

\(x^3-11x^2+30x=0\)

\(\Rightarrow x\left(x^2-11x+30\right)=0\)

\(\Rightarrow x\left[\left(x^2-10x+25\right)-\left(x-5\right)\right]=0\)

\(\Rightarrow x\left[\left(x-5\right)^2-\left(x-5\right)\right]=0\)

\(\Rightarrow x\left(5-x\right)\left(x-5-1\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\\x=6\end{array}\right.\)

\(x^3-11x^2+30x=0\)

\(\Rightarrow x\left(x^2-11x+30\right)=0\)

\(\Rightarrow x\left[\left(x^2-10x+25\right)-\left(x-5\right)\right]=0\)

\(\Rightarrow x\left[\left(x-5\right)^2-\left(x-5\right)\right]=0\)

\(\Rightarrow x\left(5-x\right)\left(x-5-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\\x-5-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=5\\x=6\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=0\\x=5\\x=6\end{matrix}\right.\)

a) \(8x^2+30x+7=0\)

\(\Leftrightarrow8\left(x^2+\frac{15}{4}x+7\right)=0\)

\(\Leftrightarrow x^2+\frac{1}{4}x+\frac{7}{2}x+\frac{7}{8}=0\) 

\(\Leftrightarrow x\left(x+\frac{1}{4}\right)+\frac{7}{2}\left(x+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{4}\right)\left(x+\frac{7}{2}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{1}{4}=0\\x+\frac{7}{2}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{7}{2}\end{array}\right.\)

b)\(x^3-11x^2+30x=0\)

\(\Leftrightarrow x\left(x^2-11x+30\right)=0\)

\(\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)

\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\\x-6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\\x=6\end{array}\right.\)

a) \(\Delta=\left(30\right)^2-4.8.7=676>0\) ( PTC2NPB )

\(X_1=\frac{-30+\sqrt{676}}{16}\)

\(X_2=\frac{-30-\sqrt{676}}{16}\)

\(=x^3-5x^2-6x^2+30x\)

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

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\\x=6\end{array}\right.\)

x³ - 11x² + 30x = 0

x(x² - 11x + 30) = 0

x(x² - 5x - 6x + 30) = 0

x[x(x - 5) - 6(x - 5)] = 0

x(x - 5)(x - 6) = 0

\(\left[\begin{array}{nghiempt}x=0\\x-5=0\\x-6=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=5\\x=6\end{array}\right.\)

\(x^2+3x-18=0\)

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

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

\(8x^2+30x+7=0\)

\(\Leftrightarrow\left(x+\frac{1}{4}\right)\left(x+\frac{7}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=0\\x+\frac{7}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=-\frac{7}{2}\end{cases}}}\)

\(x^3-11x^2+30x=0\)

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

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

\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x-2\right)\left(x+6\right)\)

b) 8x² + 30x + 7  = 0

8x² + 16x + 14x + 7  = 0

8x.(x+2) + 7.(x+2) = 0

(x+2).(8x+7) = 0

..

bn tự làm tiếp nhé! ^-^

c) x³ - 11x² + 30x = 0

x.(x² - 11x +30) = 0

\(x.\left(x^2-5x-6x+30\right)=0.\)

x.[ x.(x-5) - 6.(x-5) ] = 0

x.(x-5).(x-6) = 0

...

a/ x² + 3x - 18 = 0

x² -3x + 6x - 18 = 0

x(x-3) + 6(x-3) = 0

(x-3)(x+6) = 0

Suy ra: x - 3 = 0 hoặc x + 6 = 0

hay x = 3 hoặc x = - 6

Vậy x thuộc {3;-6}.

b/ 8x² + 30x + 7 = 0

8x² + 2x + 28x + 7 = 0

2x(4x+1) + 7(4x+1) = 0

(4x+1)(2x+7) = 0

Suy ra: 4x + 1 = 0 hoặc 2x + 7 = 0

hay x = -1/4 hoặc x = -7/2

Vậy x thuộc {-1/4; -7/2}.

c/ x³ - 11x² + 30x = 0

x(x² - 11x + 30) = 0

x(x² - 5x - 6x + 30) = 0

x.[x(x-5) - 6(x-5)] = 0

x(x-5)(x-6) = 0

Suy ra: x = 0; x - 5 = 0 hoặc x - 6 = 0

hay x = 0; x =5; x =6

Vậy x thuộc {0;5;6}.

\(x^3-11x^2+30x=0\)

\(\left(x-6\right).\left(x-5\right).x=0\)

\(=>\orbr{\begin{cases}x-6=0\\x-5=0,x=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=6\\x=5,x=0\end{cases}}\)

P/S: mk mới lớp 7 sai sót mong bỏ qua 

\(8x^2+30x+7=0\)

\(8x^2+28x+2x+7=0\)

\(2x.\left(4x+1\right)+7.\left(4x+1\right)=0\)

\(\left(2x+7\right).\left(4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=-7\\4x=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)

vậy ....

P/S sorry mk làm hơi lâu :)__chờ tí làm câu a cho

\(x^2+3x-18=0\)

\(x^2-3x+6x-18=0\)

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

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

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

P/S:nếu cách làm sai sót hay dài thì bn thồn cảm nha mk kiểm tra kết quả thấy ko sai 

x³ - 11x² + 30x = x(x² - 11x + 30) = x(x² - 6x - 5x + 30) = x[x(x - 6) - 5(x - 6)] = x(x - 5)(x - 6)

Trả lời:

x³ - 11x² + 30x

= x ( x² - 11x + 30 )

= x ( x² - 5x - 6x + 30 ) 

= x [ ( x² - 5x ) - ( 6x - 30 ) ] 

= x [ x ( x - 5 ) - 6 ( x - 5 ) ]

= x ( x - 6 ) ( x - 5 )

\(x^3-11x^2+30x\)

\(\rightarrow x.\left(x^2-11x+30\right)\)

\(\rightarrow x.\left(x^2-6x-5+30\right)\)

\(\rightarrow x.[x.\left(x-6\right)-5.\left(x-6\right)]\)

\(\rightarrow x.\left(x-5\right).x-6\)