Tìm x biết : \(|2x-3|=3-x\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

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: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)
a) \(\Leftrightarrow x^3-8-x^3+9x=1\)
\(\Leftrightarrow9x=9\Leftrightarrow x=1\)
b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)
\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

3:
a: 3^x*3=243
=>3^x=81
=>x=4
b; 2^x*16^2=1024
=>2^x=4
=>x=2
c: 64*4^x=16^8
=>4^x=4^16/4^3=4^13
=>x=13
d: 2^x=16
=>2^x=2^4
=>x=4
\(\left|2x-3\right|=3-x\)
\(\Rightarrow\orbr{\begin{cases}2x-3=3-x\\2x-3=x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+x=3+3\\2x-x=3-3\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
Việt Hoàng làm thiếu, cô mk bảo thiếu cái này là sai !
\(\left|2x-3\right|=3-x\)\(\left(đk:3-x\ge0\Leftrightarrow x\le3\right)\)
Khi đó ta có: \(\orbr{\begin{cases}2x-3=3-x\\2x-3=x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+x=3+3\\2x-x=3-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=6\\x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)( thỏa mãn )