Tìm x biết:
| x -2 | + | 3 - 2x | = 2x + 1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)
a) \(\Leftrightarrow x^3-8-x^3+9x=1\)
\(\Leftrightarrow9x=9\Leftrightarrow x=1\)
b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)
\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow2x^2+3x+1-2x^2-x+3=0\)
=>2x=-4
hay x=-2
Đặt x2 + 3x + 3 = a ; x2 - x - 1 = b ; -2x2 - 2x - 1 = c ; -1 = d
Ta nhận thấy a3 + b3 + c3 + d3 = 0 (1)
và a + b + c + d = 0
Khi đó ta có (1) <=> (a + b)3 + (c + d)3 - 3ab(a + b) - 3cd(c + d) = 0
<=> ab(a + b) + cd(c + d) = 0
<=> (a + b)(ab - cd) = 0
<=> \(\left[{}\begin{matrix}a=-b\\ab=cd\end{matrix}\right.\)
Với a = -b ta được x2 + 3x + 3 = -x2 + x + 1
<=> x2 + x + 1 = 0
<=> \(\left(x+\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\)
=> Phương trình vô nghiệm
Với ab = cd
\(\Leftrightarrow\left(x^2+3x+3\right).\left(x^2-x-1\right)=2x^2+2x+1\)
\(\Leftrightarrow\) \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)-\left(4x^2+8x+4\right)=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-\left(2x+2\right)^2=0\)
\(\Leftrightarrow\left(x^2+3x+2\right).\left(x^2-x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2.\left(x-2\right).\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)
a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)
\(\Leftrightarrow2x^2-2x-2x^2=-6\)
\(\Leftrightarrow x=3\)
b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a) đk: x khác 1; \(\dfrac{3}{2}\)
\(P=\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5}{2x-3}\right]:\left(\dfrac{3-3x+2}{1-x}\right)\)
= \(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{5-3x}{1-x}\)
= \(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}.\dfrac{1-x}{-3x+5}=\dfrac{-1}{2x-3}\)
b) Có \(\left|3x-2\right|+1=5\)
<=> \(\left|3x-2\right|=4\)
<=> \(\left[{}\begin{matrix}3x-2=4< =>x=2\left(Tm\right)\\3x-2=-4< =>x=\dfrac{-2}{3}\left(Tm\right)\end{matrix}\right.\)
TH1: Thay x = 2 vào P, ta có:
P = \(\dfrac{-1}{2.2-3}=-1\)
TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:
P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)
c) Để P > 0
<=> \(\dfrac{-1}{2x-3}>0\)
<=> 2x - 3 <0
<=> x < \(\dfrac{3}{2}\) ( x khác 1)
d) P = \(\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)
<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)
<=> 2x - 3 = x2 - 6
<=> x2 - 2x - 3 = 0
<=> (x-3)(x+1) = 0
<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)
\(|x-2|+|3-2x|=2x+1\left(1\right)\)
Ta có: \(x-2=0\Leftrightarrow x=2\)
\(3-2x=0\Leftrightarrow x=\frac{3}{2}\)
Lập bảng xét dấu:
+) Với\(x\le\frac{3}{2}\Rightarrow\hept{\begin{cases}x-2< 0\\3-2x\ge0\end{cases}\Rightarrow\hept{\begin{cases}|x-2|=2-x\\|3-2x|=3-2x\end{cases}\left(2\right)}}\)
Thay (2) vào (1) ta được:
\(2-x+3-2x=2x+1\)
\(\Leftrightarrow5-3x=2x+1\)
\(\Leftrightarrow-3x-2x=1-5\)
\(\Leftrightarrow-5x=4\)
\(\Leftrightarrow x=\frac{-4}{5}\)( chọn )
+) Với\(\frac{3}{2}< x< 2\Rightarrow\hept{\begin{cases}x-2< 0\\3-2x< 0\end{cases}\Rightarrow\hept{\begin{cases}|x-2|=2-x\\|3-2x|=2x-3\end{cases}\left(3\right)}}\)
Thay (3) vào (1) ta được:
\(2-x+2x-3=2x+1\)
\(\Leftrightarrow x-1=2x+1\)
\(\Leftrightarrow x-2x=1+1\)
\(\Leftrightarrow x=-2\)( loại)
+) Với\(x\ge2\Rightarrow\hept{\begin{cases}x-2\ge0\\3-2x< 0\end{cases}\Rightarrow\hept{\begin{cases}|x-2|=x-2\\|3-2x|=2x-3\end{cases}\left(4\right)}}\)
Thay (4) vào (1) ta được:
\(x-2+2x-3=2x+1\)
\(\Leftrightarrow3x-5=2x+1\)
\(\Leftrightarrow3x-2x=1+5\)
\(\Leftrightarrow x=6\)(chọn)
Vậy \(x\left\{\frac{-4}{5};6\right\}\)
dòng cuối là \(x\in\left\{\frac{-4}{5};6\right\}\)