Question

[6 Phạm Thế Anh]

tìm x:

|x² + |x + 3|| = x² + 2020

Sắp nộp rồi, nhanh lên

Answer 1

[No name]

) ĐKXĐ: x∉{3;−3;−2}x∉{3;−3;−2}

Ta có: P=(2x−1x+3−x3−x−3−10xx2−9):x+2x−3P=(2x−1x+3−x3−x−3−10xx2−9):x+2x−3

=((2x−1)(x−3)(x+3)(x−3)+x(x+3)(x−3)(x+3)−3−10x(x−3)(x+3)):x+2x−3=((2x−1)(x−3)(x+3)(x−3)+x(x+3)(x−3)(x+3)−3−10x(x−3)(x+3)):x+2x−3

=2x2−6x−x+3+x2+3x−3+10x(x−3)(x+3):x+2x−3=2x2−6x−x+3+x2+3x−3+10x(x−3)(x+3):x+2x−3

=3x2+6x(x−3)(x+3):x+2x−3=3x2+6x(x−3)(x+3):x+2x−3

=3x(x+2)(x−3)(x+3)⋅x−3x+2=3x(x+2)(x−3)(x+3)⋅x−3x+2

=3xx+3=3xx+3

b) Ta có: x2−7x+12=0x2−7x+12=0

⇔x2−3x−4x+12=0⇔x2−3x−4x+12=0

⇔x(x−3)−4(x−3)=0⇔x(x−3)−4(x−3)=0

⇔(x−3)(x−4)=0⇔(x−3)(x−4)=0

⇔[x−3=0x−4=0⇔[x=3(loại)x=4(nhận)⇔[x−3=0x−4=0⇔[x=3(loại)x=4(nhận)

Thay x=4 vào biểu thức P=3xx+3P=3xx+3, ta được: 

P=3⋅44+3=127P=3⋅44+3=127

Vậy: Khi x2−7x+12=0x2−7x+12=0 thì P=127