$(2x+1)^3=(2x+1)^4$
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16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
TA
21 tháng 7 2021
1) `2x(3x-1)-(2x+1)(x-3)`
`=6x^2-2x-2x^2+6x-x+3`
`=4x^2+3x+3`
2) `3(x^2-3x)-(4x+2)(x-1)`
`=3x^2-9x-4x^2+4x-2x+2`
`=-x^2-7x+2`
3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`
`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`
`=3x^2-15x-x^2+4x-4-4x^2+9`
`=-2x^2-11x+5`
4) `(2x-3)^2+(2x-1)(x+4)`
`=4x^2-12x+9+2x^2+8x-x-4`
`=6x^2-5x+5`
11 tháng 8 2021
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
30 tháng 1 2021
ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};-1;\dfrac{-3}{2};-2\right\}\)
Ta có: \(\dfrac{4}{2x+1}-\dfrac{2}{2x+3}=\dfrac{1}{2x+4}-\dfrac{3}{2x+2}\)
\(\Leftrightarrow\dfrac{4\left(2x+3\right)}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{2\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2}{\left(2x+2\right)\left(2x+4\right)}-\dfrac{3\left(2x+4\right)}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{8x+12-4x-2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2-6x-12}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}=0\)
\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{4x+10}{\left(2x+2\right)\left(2x+4\right)}=0\)
\(\Leftrightarrow\left(4x+10\right)\left(\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{1}{\left(2x+2\right)\left(2x+4\right)}\right)=0\)
\(\Leftrightarrow2\left(2x+5\right)\left(\dfrac{\left(2x+2\right)\left(2x+4\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}+\dfrac{\left(2x+1\right)\left(2x+3\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(4x^2+8x+4x+8+4x^2+6x+2x+6\right)=0\)(Vì \(\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left(2x+5\right)\left(8x^2+20x+14\right)=0\)
mà \(8x^2+20x+14>0\forall x\)
nên 2x+5=0
\(\Leftrightarrow2x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{2}\)
Vậy: \(S=\left\{-\dfrac{5}{2}\right\}\)
MB
0
1 tháng 7 2015
tách 2,3 câu ra làm 1 câu hỏi đi. bạn đăng cả đóng thế này k ai tl cho đâu. khi nào tách thì gửi link mình tl cho
20 tháng 1 2021
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
(2x +1)3 = (2x +1)4
(2x + 1)4 - (2x + 1 )3 =0
(2x + 1)3(2x +1-1) =0
2x(2x+1)3= 0
\(\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)