K
Khách

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

d) đề là gì bn

(2x+3)(4x2−6x+9)−2(4x3−1)

(4x1)3(4x3)(16x2+3)(4x−1)3−(4x−3)(16x2+3)

=64x348x2+12x1(64x3+12x48x29)=64x3−48x2+12x−1−(64x3+12x−48x2−9)

=64x348x2+12x164x312x+48x2+9=64x3−48x2+12x−1−64x3−12x+48x2+9

=8

29 tháng 2 2020

đề không rõ nên mình làm như này:

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

\(=2x^2+x-x^3-2x^2+x^3-x+3\)

\(=3\)

d) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2\)

\(=29\)

29 tháng 2 2020

\(c, C=x(2x+1)-x^2(x+2)+x^3-x+3\)

\(C=2x^2+x-x^3-2x^2+x^3-x+3\)

\(C=3\)

\(d, (2x+3)(4x^2-6x+9)-2(4x^3-1)\)

\(=(8x^3+27)-2(4x^3-1)\)

\(=8x^3+27-8x^3+2\)\(=29\)

\(e, (4x-1)^3-(4x-3)(16x^2+3)\)

\(=(64x^3-48x^2+12x-1)-(64x^3+12x-48x^2-9)\)

\(=64x^3-48x^2+12x-1-64x^3-12x+48x^2+9\)

\(=8\)

\(f, (x+1)^3-(x-1)^3-6(x+1)(x-1)\)

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

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

\(=8\)

7 tháng 12 2019

d) \(\frac{4x^2-12x+9}{9-4x^2}=-\frac{\left(2x+3\right)^2}{\left(2x-3\right)\left(2x+3\right)}=\frac{2x+3}{2x-3}\)

13 tháng 3

c)C=x(2x+1)-x^2(x+2)+x^3-x+3

=2x^2+x-x^3-2x^2+x^3-x+3

=3(không PT vào biến x)

 

31 tháng 7 2023

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

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

\(=9x^2+18x-27-x^3-2x^2+3x\)

\(=-x^3+7x^2+21x-27\)

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

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

\(=-x^3-x^2-x+3x^2+3x+3\)

\(=-x^2+2x^2+2x+3\)

o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)

\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)

\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)

\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)

q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)

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

\(=6x^3-12x^2-18x+x^2-2x-3\)

\(=6x^3-11x^2-20x-3\)

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

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

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

\(=-2x^3-7x^2-x+1\)

u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)

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

\(=-2x^3+2x^2+12x+3x^2-3x-18\)

\(=-2x^3+5x^2+9x-18\)

s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)

\(=-4x\left(x^2+3x-2\right)+5\left(x^2+3x-2\right)\)

\(=-4x^3-12x^2+8x+5x^2+15x-10\)

\(=-4x^3-7x^2+23x-10\)

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)

\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)

\(=-x^2-3+2x^4+6x+18-12x^3\)

\(=2x^4-12x^3-x^2+6x+15\)

p: (-x+9)(x^2+2x-3)

=-x^3-2x^2+3x+9x^2+18x-27

=-x^3+7x^2+21x-27

n: (-x+3)(x^2+x+1)

=-x^3-x^2-x+3x^2+3x+3

=-x^3+2x^2+2x+3

o: (-6x+1/2)(x^2-4x+2)

=-6x^3+24x^2-12x+1/2x^2-2x+1

=-64x^3+49/2x^2-14x+1

q: (6x+1)(x^2-2x-3)

=6x^3-12x^2-18x+x^2-2x-3

=6x^3-11x^2-20x-3

r: (2x+1)(-x^2-3x+1)

=-2x^3-6x^2+2x-x^2-3x+1

=-2x^3-7x^2-x+1

u: =-2x^3+2x^2+12x+3x^2-3x-18

=-2x^3+5x^2+9x-18

s: =-4x^3-12x^2+8x+5x^2+15x-10

=-4x^3-7x^2+23x-10

10 tháng 8 2017

\(\left(x-3\right)^3+\left(x+3\right)^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

\(c,x^2-4x+3=0\)

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(d,4x^2+4x+1=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0\)

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

10 tháng 8 2017

\(\left(x-3\right)^3+\left(x+3\right)^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

\(c,x^2-4x+3=0\)

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(d,4x^2+4x+1=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0\)

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

10 tháng 8 2017

\(\left(x-3\right)^3+\left(x+3\right)^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

\(b,\left(x+1\right)^3-\left(x-1\right)^3=0\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

\(c,x^2-4x+3=0\)

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(d,4x^2+4x+1=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0\)

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+2-x-3\right)\left(x+2+x+3\right)=0\)

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

7 tháng 11 2021

\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

7 tháng 11 2021

a) 9x2-49=0

(3x)2-72=0

<=> (3x-7)(3x+7)=0

th1: 3x-7=0

<=>3x=7

<=>x=\(\dfrac{7}{3}\)

th2: 3x+7=0

<=>3x=-7

<=>x=\(-\dfrac{7}{3}\)