tìm x biết 1.(x-2)^3-(x-2)(x^2+2x+4)+6(x-2)(x+2)=60 2.(5x-1)^2-(5x-4)(5x+4)=7 15/08/2021 Bởi Gabriella tìm x biết 1.(x-2)^3-(x-2)(x^2+2x+4)+6(x-2)(x+2)=60 2.(5x-1)^2-(5x-4)(5x+4)=7
Đáp án: Giải thích các bước giải: 1. $(x-2)^3 -(x-2).(x^2+2x+4)+6.(x-2).(x+2)=60$ $⇔x^3-3.x^2.2+3.x.2^2-2^3-(x^3-8)+6.(x^2-4)=60 $ $⇔ x^3-6x^2+12x-8 -x^3+8+6x^2-24=60$ $⇔12x=84$ $⇔ x=7 $ Vậy $x=7$ 2. $(5x-1)^2-(5x-4).(5x+4)=7$ $⇔(5x)^2-2.5x.1+1-[(5x)^2-4^2]=7$ $⇔25x^2-10x+1-(25x^2-16)=7$ $⇔25x^2-10x+1-25x^2+16=7$ $⇔-10x=-10$ $⇔ x=1$ Vậy $x=1$ Bình luận
Đáp án: $1. (x – 2)³ – (x – 2)(x² + 2x + 4) + 6(x – 2)( x + 2) = 60$ $<=> x³ – 6x² + 12x – 8 – ( x³ – 8) + 6(x² – 4) = 60$ $<=> x³ – 6x² + 12x – 8 – x³ + 8 + 6x² – 24 = 60$ $<=> 12x – 24 = 60$ $<=> 12x = 84$ $<=> x = 7$ Vậy $x = 7$ $2. ( 5x – 1)² – ( 5x – 4)(5x + 4) = 7$ $<=> 25x² – 10x + 1 – 25x² + 16 = 7$ $<=> – 10x + 17 = 7$ $<=> – 10x = -10$ $<=> x = 1 $ Vậy $x = 1$ Bình luận
Đáp án:
Giải thích các bước giải:
1.
$(x-2)^3 -(x-2).(x^2+2x+4)+6.(x-2).(x+2)=60$
$⇔x^3-3.x^2.2+3.x.2^2-2^3-(x^3-8)+6.(x^2-4)=60 $
$⇔ x^3-6x^2+12x-8 -x^3+8+6x^2-24=60$
$⇔12x=84$
$⇔ x=7 $
Vậy $x=7$
2.
$(5x-1)^2-(5x-4).(5x+4)=7$
$⇔(5x)^2-2.5x.1+1-[(5x)^2-4^2]=7$
$⇔25x^2-10x+1-(25x^2-16)=7$
$⇔25x^2-10x+1-25x^2+16=7$
$⇔-10x=-10$
$⇔ x=1$
Vậy $x=1$
Đáp án:
$1. (x – 2)³ – (x – 2)(x² + 2x + 4) + 6(x – 2)( x + 2) = 60$
$<=> x³ – 6x² + 12x – 8 – ( x³ – 8) + 6(x² – 4) = 60$
$<=> x³ – 6x² + 12x – 8 – x³ + 8 + 6x² – 24 = 60$
$<=> 12x – 24 = 60$
$<=> 12x = 84$
$<=> x = 7$
Vậy $x = 7$
$2. ( 5x – 1)² – ( 5x – 4)(5x + 4) = 7$
$<=> 25x² – 10x + 1 – 25x² + 16 = 7$
$<=> – 10x + 17 = 7$
$<=> – 10x = -10$
$<=> x = 1 $
Vậy $x = 1$