Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cU-----------------------------------------'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''l;;;;;;;;;;;;;;g'''''''''''''''''b'''''''''''''''''''''''mgcxryjwrsmđxibkvkkkkkii111111hỳdc cccccgccccffffffffftsydfughjkmn oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooffffffffffffffffffffff ưooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
dsfazzoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooool
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
Rút gọn biểu thức
a).x(2x2-3)-x2(5x+1)+x2
b).3x(x-2)-5x(1-x)-8(x2-3)
c).1/2x2(6x-3)-x(x2+1/2)+1/2(x+4)
a) x(2x^2 -3) -x^2 (5x+1 ) + x^2
<=> 2x^3 -3x -5x^3 -x^2 +x^2
<=>3x^3 -3x
b) 3x(x-2) -5x(1-x)-8(x^2 -3)
=3x^2 -6x -5x +5x^2 -8x^2 +24
= -11x+24
Bài 1
a) (6x4y2 - 3x3y3) : 3x3y2 = 6x4y2 : 3x3y2 - 3x3y3 : 3x3y2 = 2x - y
b) (2x - 1)(x2 - x + 3) = 2x3 - 2x2 + 6x - x2 + x - 3 = 2x3 - 3x2 + 7x - 3
Bài 2
1) (x - 2)2 - (x - 3)2 = (x - 2 - x + 3)(x - 2 + x - 3) = 2x - 5>
2) 4x2 - 4xy + 2y2 + 1 = (4x2 - 4xy + y2) + y2 + 1 = (2x - y)2 + y2 + 1 > 0
vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)
\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)
\(=6x^2+12x+x+2-6x^2+10x\)
\(=23x+2\)
a) (6x + 1)(x + 2) - 2x(3x - 5)
= 6x2 + 12x + x + 2 - 6x2 + 10x
= (6x2 - 6x2) + (12x + x + 10x) + 2
= 23x + 2
b) (2x - 1)2 - (2x - 3)(2x + 3)
= 4x2 - 4x + 1 - 4x2 + 9
= (4x2 - 4x2) - 4x + (1 + 9)
= -4x + 10
c) (2x - 3)3 - (3x + 1)(5 - 4x) - 16x2
= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2
= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5
= 8x3 - 40x2 + 43x - 5
d) (3x + 2) - (x - 5) - x(3x - 13)
= 3x + 2 - x + 5 - 3x2 + 13x
= (3x - x + 13x) + (2 + 5) - 3x2
= 15x + 7 - 3x2
\(1.a,Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}=\frac{x+3}{2x+1}+\frac{7-x}{2x+1}\)
\(=\frac{x+3+7-x}{2x+1}=\frac{10}{2x+1}\)
\(b,\) Vì \(x\inℤ\Rightarrow\left(2x+1\right)\inℤ\)
Q nhận giá trị nguyên \(\Leftrightarrow\frac{10}{2x+1}\) nhận giá trị nguyên
\(\Leftrightarrow10⋮2x+1\)
\(\Leftrightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(\left(2x+1\right):2\) dư 1 nên \(2x+1=\pm1;\pm5\)
\(\Rightarrow x=-1;0;-3;2\)
Vậy.......................
Bài 1:
a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10
b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61
Bài 2:
a)(2x-1)^2-(x+3)^2 = 0
<=> (2x-1-x-3).(2x-1+x+3) =0
<=>(x-4).(3x+2) = 0
<=> x-4 = 0 hoặc 3x+2=0
*x-4=0 => x=4
*3x+2 = 0 => 3x=-2 => x=-2/3
b)x^2(x-3)+12-4x=0 <=> x^2(x-3) - 4(x-3) =0 <=> (x-3).(x-2)(x+2) <=> x-3=0 hoặc x-2=0 hoặc x+2 =0
*x-3=0 => x=3
*x-2=0 =>x=2
*x+2=0 =>x=-2
c) 6x^3 -24x =0 <=> 6x(x^2 -4)=0 <=> 6x(x-2)(x+2)=0 <=> x=0 hoặc x-2 =0 hoặc x+2=0 <=> x=0 hoặc x=2 hoặc x=-2