b1Tìm x 1. (4x-3).(4x+3)-(4x+1)^2 =3.(2x-5) 2. 2x(9+2x)-(2x-5)^2=15 b2 Rút gọn B = 3x^3-9x^2/2x^3-12x^2+18x 03/11/2021 Bởi Maria b1Tìm x 1. (4x-3).(4x+3)-(4x+1)^2 =3.(2x-5) 2. 2x(9+2x)-(2x-5)^2=15 b2 Rút gọn B = 3x^3-9x^2/2x^3-12x^2+18x
(4x-3).(4x+3)-(4x+1)^2 =3.(2x-5) =>16x^2-9-(16x^2+8x+1)=6x-15 =>16x^2-9-16x^2-8x-1-6x+15=0 =>-14x+5=0 =>x=5/14 2x(9+2x)-(2x-5)^2=15 =>18x+4x^2-4x^2+20x=15+25 =>38x=40 =>19x=20 =>x=20/19 B = 3x^3-9x^2/2x^3-12x^2+18x =3x^2(x-3)/2x(x^2-6x+9) =3x(x-3)/2(x-3)^2 =3x/2(x-3) Bình luận
Đáp án+giải thích các bước giải: B1: 1, $(4x-3).(4x+3)-(4x+1)^2=3.(2x-5)$ $16x^2-9-16x^2-8x-1=6x-15$ $16x^2-16x^2-8x-6x=-15+1+9$ $-14x=-5$ $x=\dfrac{5}{14}$ $\text{Vậy x=}$ $\dfrac{5}{14}$ 2, $2x(9+2x)-(2x-5)^2=15$ $18x+4x^2-4x^2+20x-25=15$ $18x+4x^2-4x^2+20x=15+25$ $38x=40$ $x=\dfrac{20}{19}$ $\text{Vậy x=}$ $\dfrac{20}{19}$ B2: $B=\dfrac{ 3x^3-9x^2}{2x^3-12x^2+18x}$ $=\dfrac{3x^2(x-3)}{2x(x^2-6x+9)}$ $=\dfrac{3x^2(x-3)}{2x(x-3)^2}$ $=\dfrac{3x}{2(x-3)}$ Bình luận
(4x-3).(4x+3)-(4x+1)^2 =3.(2x-5)
=>16x^2-9-(16x^2+8x+1)=6x-15
=>16x^2-9-16x^2-8x-1-6x+15=0
=>-14x+5=0
=>x=5/14
2x(9+2x)-(2x-5)^2=15
=>18x+4x^2-4x^2+20x=15+25
=>38x=40
=>19x=20
=>x=20/19
B = 3x^3-9x^2/2x^3-12x^2+18x
=3x^2(x-3)/2x(x^2-6x+9)
=3x(x-3)/2(x-3)^2
=3x/2(x-3)
Đáp án+giải thích các bước giải:
B1:
1,
$(4x-3).(4x+3)-(4x+1)^2=3.(2x-5)$
$16x^2-9-16x^2-8x-1=6x-15$
$16x^2-16x^2-8x-6x=-15+1+9$
$-14x=-5$
$x=\dfrac{5}{14}$
$\text{Vậy x=}$ $\dfrac{5}{14}$
2,
$2x(9+2x)-(2x-5)^2=15$
$18x+4x^2-4x^2+20x-25=15$
$18x+4x^2-4x^2+20x=15+25$
$38x=40$
$x=\dfrac{20}{19}$
$\text{Vậy x=}$ $\dfrac{20}{19}$
B2:
$B=\dfrac{ 3x^3-9x^2}{2x^3-12x^2+18x}$
$=\dfrac{3x^2(x-3)}{2x(x^2-6x+9)}$
$=\dfrac{3x^2(x-3)}{2x(x-3)^2}$
$=\dfrac{3x}{2(x-3)}$