Question

Phương pháp tách hạng tử

a) \(x^2-6x+8\)

b) \(x^2-4x+3\)

c) \(x^2-x-12\)

Answer 1

a) \(x^2-6x+8=x^2-4x-2x+8=x\left(x-4\right)-2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)

b) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)

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

Answer 2

a) x^2 -6x +8 

= x^2 - 2x -4x +8

= x(x -2) - 4 ( x-2)

= (x-2)(x-4)

b) x^2 - 4x +3

= x^2 -x - 3x +3

= x(x-1) -3(x-1)

= (x-1)(x-3)

Answer 3

c) x^2 -x-12

=x^2 - 3x + 4x -12

=x(x-3) +4(x-3)

=(x-3)(x+4)

1000% đúng

chúc bạn học tốt

Answer 4

a) \(x^2-6x+8\) \(=x^2-2x-4x+8\) \(=x.\left(x-2\right)-2\left(x-2\right)\)\(=\left(x-2\right).\left(x-2\right)=\left(x-2^2\right)\)

b) \(x^2-4x+3=x^2-x-3x+3=x.\left(x-1\right)-3.\left(x-1\right)\)\(=\left(x-1\right).\left(x-3\right)\)

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

k nha~!!!!

Answer 5

a) \(x^2-6x+8\)

\(=x^2-6x+9-1=\left(x-3\right)^2-1=\left(x-3-1\right)\left(x-3+1\right)=\left(x-4\right)\left(x-2\right)\)

b) \(x^2-4x+3\)

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

c) \(x^2-x-12\)

\(=x^2-2\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}-12\)

\(=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}=\left(x-\frac{1}{2}\right)^2-3,5^2=\left(x-\frac{1}{2}-3,5\right)\left(x-\frac{1}{2}+3,5\right)=\left(x-4\right)\left(x+3\right)\)

Answer 6

câu a phải là \(\left(x-2\right)^2\) nha!!! mik ấn nhầm...

bạn thông cảm nha!!!

Answer 7

câu a mik ấn nhầm số 2...nên két quả của mik câu a la sai nha