\(x^2-8x+15\)

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

\(=x\left(x-3\right)-5 \left(x-3\right)\)

\(=\left(x-5\right)\left(x-3\right)\)

\(x^2-8x+15\)

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

\(=x\left(x-3\right)-5\left(x-3\right)\)

\(=\left(x-3\right)\left(x-5\right)\)

Tham khảo nhé~

A)\(x\left(x-1\right)+6\left(x-3\right)\left(x+3\right)\)

\(=x^2-x+6\left(x^2-9\right)\)

\(=x^2-x+6x^2-54\)

\(=7x^2-x-54\)

F.\(\left(2-x\right)\left(2+x\right)-2x\left(x-7\right)+x\left(x+1\right)\)

\(=4-x^2-2x^2+14x+x^2+x\)

\(=-2x^2+15x+4\)

\(x^2-10x+16\)

\(=\left(x^2-2x\right)-\left(8x-16\right)\)

\(=x.\left(x-2\right)-8\left(x-2\right)\)

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

Tham khảo nhé~

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

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

\(\orbr{\Leftrightarrow\begin{cases}x=0\\x=5\end{cases}or}:x=6\)

\(\Leftrightarrow\left(x-5\right)\left(x-6\right)=0\text{:}x=0\)

Trường hợp 1 \(x-5=0\Rightarrow x=5\)

Trường hợp 2\(x-6=0\Rightarrow x=6\)

Vậy \(x\left\{5,6\right\}.\)

a) x² + 3x - 18 = 0

x² + 6x - 3x - 18 = 0

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

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

=> x + 6 = 0 => x = -6

x-3 =0 => x = 3

KL:./.

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

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

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

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

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

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

...

      \(x^2-10x+16\)

\(=x^2-2x-8x+16\)

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

      \(x^2-8x+15\)

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

\(=x\left(x-3\right)-5\left(x-3\right)=\left(x-3\right)\left(x-5\right)\)

c) x² -10x + 16

= x² - 2x - 8x + 16

= x.(x-2) - 8.(x-2)

= (x-2).(x-8)

d) x² - 8x + 15

= x² - 3x - 5x + 15

= x.(x-3) - 5.(x-3)

= (x-3).(x-5)

      \(x^2+7x+12\)

\(=x^2+3x+4x+12\)

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

      \(x^2+6x+8\)

\(=x^2+2x+4x+8\)

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

a) x² + 7x + 12

= x² + 3x + 4x + 12

= x.(x+3) + 4.(x+3)

= (x+3).(x+4)

b) x² + 6x + 8

= x² + 2x + 4x + 8 

= x.(x+2) + 4.(x+2)

= (x+2).(x+4)

Nguyen giang tham khảo bạn nhé!

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