(2x- 3 ) \(^{x+3}\)= (2x - 3) \(^{x+1}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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`
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
29 tháng 10 2023
c: \(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\)
\(=4x^2+12x+9+4x^2-12x+9-\left(4x^2-9\right)\)
\(=8x^2+18-4x^2+9=4x^2+27\)
d: \(\left(x-1\right)\cdot\left(x^2+x+1\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(=\left(x-1\right)\left(x^2+x\cdot1+1^2\right)-\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]\)
\(=x^3-1-8x^3-27=-7x^3-28\)
e: \(\left(x+1\right)^3-\left(x-1\right)^3-6x^2\)
\(=x^3+3x^2+3x+1-6x^2-\left(x^3-3x^2+3x-1\right)\)
\(=x^3-3x^2+3x+1-x^3+3x^2-3x+1\)
=2
21 tháng 10 2021
a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)
\(=6x^2+3x+2x^2+2x-3x-3\)
\(=8x^2+2x-3\)
Y
18 tháng 4 2022
\(a,\dfrac{x-3}{4}+\dfrac{2x-1}{3}=-\dfrac{x}{6}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)+4\left(2x-1\right)+2x}{12}=0\)
\(\Leftrightarrow3x-9+8x-4+2x=0\)
\(\Leftrightarrow13x-13=0\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
\(b,\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(2x-1\right)-\left(2x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-3-2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
11 tháng 7 2021
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
11 tháng 7 2021
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
\(\left(2x-3\right)^{x+3}=\left(2x-3\right)^{x+1}\)
\(\Leftrightarrow\left(2x-3\right)^{x+3}-\left(2x-3\right)^{x+1}=0\)
\(\Leftrightarrow\left(2x-3\right)^{x+1}.\left[\left(2x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-3\right)^{x+1}=0\\\left(2x-3\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x-1=\pm1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1;x=0\end{cases}}\)
Vậy \(x\in\left\{0;1;\frac{3}{2}\right\}\)