\((2x+3)(4x^2-6x+9)-8x(x^2-2)\)

">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

\(\left(2x+3\right)\left(4x^2-6x+9\right)-8x\left(x^2-2\right)\)

\(=\left(2x+3\right)\left[\left(2x^2\right)-2x.3^2\right]-8x\left(x^2-2\right)\)

\(=\left(2x\right)^3+3^3-8x^3+16x\)

\(=18x^3+27-8x^3+16x\)

\(=16x+27\)

24 tháng 8 2021

(2x + 3)(4x2 - 6x + 9) - 8x(x2 - 2)

= (2x)3 + 33 - 8x(x2 - 2)

= 8x3 + 9  - 8x3 + 16x

= 9 + 16x

 Chúc bạn học tốt

2 tháng 1 2020

\(K=\left(x^2y-3\right)^2-\left(2x-y\right)^3+xy^2\left(6-x^3\right)+8x^3-6x^2y-y^3\)

\(=x^4y^2-6x^2y+9-4x^2+4xy-y^2+6xy^2-x^4y^2+8x^3-6x^2y-y^3\)

\(=-12x^2y+9-4x^2+4xy-y^2+6xy^2+8x^3-y^3\)

6 tháng 1 2018

https://olm.vn/hoi-dap/question/1027904.html

tk nhé 

^_^

6 tháng 1 2018

\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{ }\)

\(P=\frac{x^4\left(2x-1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(P=\frac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2}{2x+1}\)

\(P=\frac{x^4-1}{2x+1}+\frac{2}{2x+1}\)

\(P=\frac{x^4+1}{2x+1}\)

Vậy \(P=\frac{x^4+1}{2x+1}\)

17 tháng 8 2020

1) \(8x^3+12x^2+6x+1=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\)

\(=\left(2x+1\right)^3=\left(2.-2+1\right)^3=-27\)

2) \(8x^3-12x+6x-1=\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3\)

\(=\left(2x-1\right)^3=\left(2.-\frac{1}{2}-1\right)^3=-8\)

3)\(\left(1-2x\right)^2-\left(3x+1\right)^2=\left(1-2x+3x+1\right)\left(1-2x-3x-1\right)\)

\(=\left(x+2\right)\left(-5x\right)=\left(-2+2\right).\left(-5.-2\right)=0\)

4) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x-3y\right)\left[\left(2x\right)^2+2x.3y+\left(3y\right)^2\right]\)

\(=\left(2x\right)^3-\left(3y\right)^3=\left(2.-\frac{1}{2}\right)^3-\left(3.-\frac{1}{3}\right)^3=-1-\left(-1\right)=0\)

17 tháng 8 2020

1) Ta có : \(8x^3+12x^2+6x+1\)

\(=\left(2x+1\right)^3=\left(2.-2+1\right)^3=\left(-3\right)^3=-27\)

b) \(8x^3-12x^2+6x-1\)

\(=\left(2x-1\right)^3=\left[2.\left(-\frac{1}{2}\right)-1\right]^3=-8\)

9 tháng 1 2017

a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)

17 tháng 1 2021

Câub mô

 

NV
14 tháng 11 2019

ĐKXĐ; ...

a/ \(P=\frac{x^2}{x+4}\left[\frac{\left(x+4\right)^2}{x}\right]+9=x\left(x+4\right)+9=\left(x+2\right)^2+5\ge5\)

\(P_{min}=5\) khi \(x=-2\)

b/ \(Q=\left(\frac{\left(x+2\right)\left(x^2-2x+4\right).4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)\left(x-2\right)\left(x+2\right)}-\frac{4x}{x-2}\right).\frac{x\left(x-2\right)^3}{-16}\)

\(=\left(\frac{4\left(x^2-2x+4\right)-4x\left(x-2\right)}{\left(x-2\right)^2}\right).\frac{-x\left(x-2\right)^3}{16}\)

\(=\frac{16}{\left(x-2\right)^2}.\frac{-x\left(x-2\right)^3}{16}=-x\left(x-2\right)=-x^2+2x\)

\(=1-\left(x-1\right)^2\le1\)

\(Q_{max}=1\) khi \(x=1\)

27 tháng 6 2018

\(x^2+6x+9=\left(x+3\right)^2\)

--

\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)

--

\(x^3+12x^2+48x+64=\left(x+4\right)^3\)

28 tháng 6 2018

1) \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)

\(=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)

\(=\dfrac{2x^2+50}{x^2+25}\)

\(=\dfrac{2\left(x^2+25\right)}{x^2+25}=2\)

2) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(=x^3+3^3-54-x^3\)

\(=27-54=-27\)

3) \(\left(2x+y\right)^2-\left(y+3x\right)^2\)

\(=4x^2+4xy+y^2-y^2-6xy-9x^2\)

\(=-5x^2-2xy\)

4) \(\left(2x+1\right)^3-\left(2x-1\right)^3-24x^2\)

\(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2\)

\(=2\)