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a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)
b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)
c: \(=\dfrac{x}{2x+1}-\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}-\dfrac{\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x-1-\left(x-2\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x^2-x-1-2x^2-x+4x+2}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{1}{2x-1}\)
d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)
\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)

a) 2(x-1)2 - 4(x+3)2 + 2x(x-5)
= 2(x2 -2x +1)- 4(x2 + 6x +9) + 2x2 -10x
= 2x2 - 4x + 2 -4x2 - 24x - 36 + 2x2 - 10x
= (2x2 + 2x2 - 4x2) - (4x + 24x+10x) +(2-36)
= -38x-34
b) 2(2x+5)2 -3(4x+1)(1-4x)
= 2(4x2 + 20x + 25) + 3(4x+1)(4x-1)
= 8x2 +40x + 50 + 3(16x2 -1)
= 8x2 + 40x + 50 + 48x2 - 3
=56x2 +40x + 47
a, \(2\left(x-1\right)^2-4\left(x+3\right)^2+2x\left(x-5\right)\)
\(=2\left(x^2-2x+1\right)-4\left(x^2+6x+9\right)+2x\left(x-5\right)\)
\(=2x^2-4x+2-4x^2-24x-36+2x^2-10=-28x-44\)
b, \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)
\(=8x^2+40x+50-3+48x^2=56x^2+40x+47\)

1.
a, (x-1) . (x+1) . (x+2) = (x2 - 1)(x + 2) = x3 - x + 2x2 - 2
b, \(\frac{1}{2x^2y^2}\). (2x+y) . (2x-y) = \(\frac{1}{2x^2y^2}\).[(2x)2 - y2] =\(\frac{1}{2x^2y^2}\)(4x2 - y2) = \(\frac{4x^2-y^2}{2x^2y^2}\)
c, (x-1/2). (x+1/2) . (4x-1) = \(\left(x^2-\frac{1}{4}\right)\left(4x-1\right)=4x^3-x-x^2+\frac{1}{4}\)
sửa lại câu a cho hokage naruto
a) ( x - 1 ) . ( x + 1 ) . ( x + 2 )
= x^2 + x - x - 1 . ( x + 2 )
= ( x^2 - 1 ) . ( x + 2 )
= x^3 + 2x^2 - x - 2
= x^2 + 2x^2 - 2
( chỉ góp ý câu a vậy thôi ) các câu khác ko có ý kiến
Đáp án\(\in\){Thiếu,Đề chưa đầy đủ}
\(\left(2x-1\right).\left(4x^2+2x+1\right)\)
\(=8x^3+4x+2x-4x^2-2x-1\)
\(=8x^3-4x^2+4x-1\)