tìm x biết : x.(2x-1)-(x-2).(2x+3)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54\)
\(\Leftrightarrow9x^3-18x^2+9x-29=0\)
\(\Leftrightarrow x=2,208024627\)
\(a,2^{x+1}=32\\ 2^{x+1}=2^5\\ x+1=5\\ x=4\\ b,2^{2x}+2^{2x+1}=48\\ 2^{2x}+2\cdot2^{2x}=48\\ 3\cdot2^{2x}=48\\ 2^{2x}=16\\ 2^{2x}=2^4\\ 2x=4\\ x=2\)
\(c,3^x+5\cdot3^{x+1}=144\\ 3^x+15\cdot3^x=144\\ 16\cdot3^x=144\\ 3^x=9\\ 3^x=3^2\\ x=2\\ d,3^{x+5}=9^{x+1}\\ 3^{x+5}=3^{2x+2}\\ x+5=2x+2\\ x=3\)
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
a: \(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}+\dfrac{4x^2}{x^2-9}\right):\dfrac{2x+1-x-3}{x+3}\)
\(=\dfrac{-x^2-6x-9+x^2-6x+9+4x^2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-2}\)
\(=\dfrac{4x^2-12x}{x-3}\cdot\dfrac{1}{x-2}=\dfrac{4x}{x-2}\)
b: \(2x^2-5x+2=0\)
=>(x-2)(2x-1)=0
=>x=1/2
Thay x=1/2 vào P, ta được:
\(P=\left(4\cdot\dfrac{1}{2}\right):\left(\dfrac{1}{2}-2\right)=2:\dfrac{-3}{2}=\dfrac{-4}{3}\)
(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0
=> 2015.2x + (1 + 2 + 3 + ... + 2015) = 0
=> 4030x + (2015 + 1).2015 : 2 = 0
=> 4030x = -2031120
=> x = -504
2x2 - x - 2x2 - 3x + 4x + 6
x = 6