(2x + 1 - 1/1-2x):(2x- 4x2/2x-1)
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CM
3 tháng 10 2018
Ta có:
8 x 3 - 1 = 2 x 3 - 1 3 = 2 x - 1 . 4 x 2 + 2 x + 1 = - 1 - 2 x . 4 x 2 + 2 x + 1
Do đó, ( 8 x 3 - 1 : 1 - 2 x = - 4 x 2 + 2 x + 1 = - 4 x 2 - 2 x - 1
Chọn B. - 4 x 2 - 2 x - 1
2 tháng 10 2021
\(M=343-8x^3-64+8x^3=279\\ N=8x^3-1-1+8x^3=16x^3=16\cdot1000=16000\)
VC
2
29 tháng 11 2021
\(\dfrac{1-2x}{2x}+\dfrac{2x}{2x-1}+\dfrac{1}{2x-4x^2}=\dfrac{1-2x}{2x}-\dfrac{2x}{1-2x}+\dfrac{1}{2x\left(1-2x\right)}=\dfrac{\left(1-2x\right)^2}{2x\left(1-2x\right)}-\dfrac{2x.2x}{2x\left(1-2x\right)}+\dfrac{1}{2x\left(1-2x\right)}=\dfrac{1-4x+4x^2-4x^2+1}{2x\left(1-2x\right)}=\dfrac{2-4x}{2x\left(1-2x\right)}=\dfrac{2\left(1-2x\right)}{2x\left(1-2x\right)}=\dfrac{1}{x}\)
30 tháng 11 2021
\(=\dfrac{\left(2x-1\right)\left(1-2x\right)}{2x\left(2x-1\right)}+\dfrac{4x^2}{2x\left(2x-1\right)}+\dfrac{-1}{2x\left(2x-1\right)}=\dfrac{\left(2x-1\right)\left(1-2x\right)+4x^2-1}{2x\left(2x-1\right)}=\dfrac{2x-4x^2-1+2x+4x^2-1}{2x\left(2x-1\right)}=\dfrac{4x-2}{2\left(2x-1\right)}=\dfrac{2\left(2x-1\right)}{2\left(2x-1\right)}=\dfrac{2}{2}=1\)
14 tháng 10 2021
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
14 tháng 10 2021
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
28 tháng 7 2021
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
28 tháng 7 2021
5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
HC
1-3x/2x + 3x-2/2x-1 + 2-3x/4x2-2x
2
TM
20 tháng 12 2021
\(2x;2x-1;4x^2-2x=2x\left(2x-1\right)\)
\(MTC=2x\left(2x-1\right)\)
\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{2-3x}{4x^2-2x}\)
\(=\dfrac{\left(1-3x\right).2x\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{\left(3x-2\right).2x}{\left(2x-1\right).2x}+\dfrac{2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{2x\left(1-3x\right)\left(2x-1\right)+2x\left(2x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{-8x^2+4x+4x^2-4x+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2-3x+2}{2x\left(2x-1\right)}\)
#AEZn8
20 tháng 12 2021
\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{2-3x}{4x^2-2x}=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}+\dfrac{2-3x}{2x\left(2x-1\right)}=\dfrac{-6x^2+5x-1}{2x\left(2x-1\right)}+\dfrac{6x^2-4x}{2x\left(2x-1\right)}+\dfrac{2-3x}{2x\left(2x-1\right)}=\dfrac{\left(-6x^2+6x^2\right)+\left(5x-4x-3x\right)+\left(-1+2\right)}{2x\left(2x-1\right)}=\dfrac{-2x}{2x\left(2x-1\right)}=\dfrac{-1}{2x-1}\)
17 tháng 9 2021
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
18 tháng 9 2021
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
10 tháng 10 2021
a (3-2x)2 = 6 - 4x
b (xy+5)2 = 2xy + 10
c (2x+1)(1-2x) = 2x - 4x2 + 1 - 2x = 4x2 + 1
d (1-5x)3 = 3-15x
e (2x+y)(4x2 - 4xy + y2) = 8x3 -8x2y+2xy2 + 4x2y-4xy2 + y3 = 8x3 + y3 - 4x2y - 2xy2