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a) (2x+1)^2+2(4x^2-2)+(2x-1)^2=4x2+4x+1+8x2-4+4x2-4x+1=16x2-2
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
a: \(=x^3-27-x^3-27x+x+27=-26x\)
b: \(=x^2-14x-10x^2+20x-10=-9x^2+6x-10\)
c: \(\Leftrightarrow2x^2-4x-4x^2-6x+2x+3=0\)
=>3=0(vô lý)
\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)
\(=6x^2+12x+x+2-6x^2+10x\)
\(=23x+2\)
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
a ) \(\left(2x-3\right)^3-x\left(2x-1\right)^2\)
\(=\left(2x\right)^3-3.\left(2x\right)^2.3+3.2x.3^2-3^3-x.\left(2x\right)^2-2.2x.1+1^2\)
\(=8x^3-3.4x^2.3+3.2x.9-27-x.4x^2-2.2x.x+1\)
\(=8x^3-36x^2+54x-27-4x^3-4x^2+1\)
\(=4x^3-40x^2+54x-26\)
a ) (2x−3)^3−x(2x−1)^2
=(2x)^3−3.(2x)^2.3+3.2x.3^2−33^−x.(2x)^2−2.2x.1+1^2
=8x^3−3.4x^2.3+3.2x.9−27−x.4x^2−2.2x.x+1
=8x^3−36x^2+54x−27−4x^3−4x2+1
=4x^3−40x^2+54x−26
a) (2x - 1)(3x + 5) - 2(-4x + 1)2 = 6x2 + 10x - 3x - 5 - 2(16x2 - 8x + 1) = 6x2 - 3x - 5 - 32x2 + 16x - 2 = -26x2 + 13x - 7
b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)
c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)
= \(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{2x-9+2x^2-3x-2-x^2+9}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{x^2-2x+x-2}{\left(x-3\right)\left(x-2\right)}\)
= \(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\frac{x+1}{x-3}\)
d) (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)
= (x - 1 - x - 1)[(x - 1)2 + (x - 1)(x + 1) + (x + 1)2] + 6(x2 - 1)
= -2(x2 - 2x + 1 + x2 - 1 + x2 + 2x + 1) + 6x2 - 6
= -2(3x2 + 1) + 6x2 - 6
= -6x2 - 2 + 6x2 - 6
= -8
e) (2x + 7)2 - (4x + 14)(2x - 8) + (8 - 2x)2
= (2x + 7)2 - 2(2x + 7)(2x - 8) + (2x - 8)2
= (2x + 7 - 2x + 8)2
= 152 = 225
a) 9x^2-6x+1+ 12x^2-2+ 4x^2-4x+1
=25x^2-10x
b) (x^2+2x+1)(x-3) -(x^3-27)
=x^3-3-x^3+27
=24
Nhớ ấn cho mfinh nhé, mình chưa có điểm cộng nào cảm ơn
\(P=\left(\dfrac{2}{x-3}+\dfrac{x^2+3}{9-x^2}+\dfrac{x-1}{x+3}\right):\left(\dfrac{2x-1}{2x+1}-1\right)\)
\(=\left(\dfrac{2}{x-3}-\dfrac{x^2+3}{x^2-9}+\dfrac{x-1}{x+3}\right):\left(\dfrac{2x-1}{2x+1}-\dfrac{2x+1}{2x+1}\right)\)
\(=\dfrac{2\left(x+3\right)-x^2-3+\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x-1-2x-1}{2x+1}\)
\(=\dfrac{2x+6-x^2-3+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{-2}{2x+1}\)
\(=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{-2}{2x+1}\)
\(=\dfrac{-2}{\left(x+3\right)}\times\dfrac{2x+1}{-2}\)
\(=\dfrac{2x+1}{x+3}\)