2x - 3 - 3*(x - 5) = 4*(3 - x) - 18
Giải giúp mình với
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
PT $\Leftrightarrow 4x^2+4x+1=3(x^2-4)+18$
$\Leftrightarrow 4x^2+4x+1=3x^2+6$
$\Leftrightarrow x^2+4x-5=0$
$\Leftrightarrow (x-1)(x+5)=0$
$\Leftrightarrow x-1=0$ hoặc $x+5=0$
$\Leftrightarrow x=1$ hoặc $x=-5$
\(\left(2x+1\right)^2=3\left(x-2\right)\left(x+2\right)+18\)
\(\Leftrightarrow4x^2+4x+1=3\left(x^2-4\right)+18\)
\(\Leftrightarrow4x^2+4x+1=3x^2-12+18\)
\(\Leftrightarrow4x^2+4x+1=3x^2+6\)
\(\Leftrightarrow4x^2-3x^2+4x=6-1\)
\(\Leftrightarrow x^2+4x=5\)
\(\Leftrightarrow x^2+4x-5=0\)
\(\Leftrightarrow x^2+5x-x-5=0\)
\(\Leftrightarrow x\left(x+5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{-5;1\right\}\)
\(\frac{3}{x-5}=-\frac{4}{x+2}\)
=> 3 ( x + 2 ) = 4 ( x - 5 )
=> 3x + 6 = 4x - 20
=> 3x - 4x = - 6 - 20
=> - 1x = - 26
=> x = 26
1) 3/x-5=-4/x+2
3x+6=-4x+20
7x . =14
x . =2
2) x+3/-4=-9/x+3
(x+3)^2=36
Ta có hai trường hợp:
*x+3=6=)x=3
*x+3=-6=)x=-9
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
1: =>(x+2018)(6x-3)=0
=>x+2018=0 hoặc 6x-3=0
=>x=1/2 hoặc x=-2018
2: x(x-11)+3(11-x)=0
=>(x-11)(x-3)=0
=>x=11 hoặc x=3
4: =>(x+5)(2x-4)=0
=>2x-4=0 hoặc x+5=0
=>x=2 hoặc x=-5
3: =>(x-3)(x+2)=0
=>x=3 hoặc x=-2
Bài 1:
\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)
\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)
\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)
Bài 2:
\(x\left(x-11\right)+3\left(11-x\right)=0\)
\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)
Câu 3:
\(x\left(x-3\right)-2\left(3-x\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Câu 4:
\(2x\left(x+5\right)-4\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a: \(A=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
\(=-x-15\)
\(=-\left(-1\right)-15=1-15=-14\)
\(2x-3-3x+3.5=4.3-4.x-18\)
<=> \(2x-3-3x+15=12-4x-18\)
<=> \(2x-3x+4x=12-18-15+3\)
<=>\(3x=-18\)
<=> x=-18/3=-6
\(2x-3-3\left(x-5\right)=4\left(3-x\right)-18\)
\(\Leftrightarrow2x-3-3x+15=12-4x-18\)
\(\Leftrightarrow\left(2x-3x\right)+\left(15-3\right)=\left(12-18\right)-4x\)
\(\Leftrightarrow-x+12=-6-4x\)
\(\Leftrightarrow12=-6-3x\)
\(\Leftrightarrow3x=-6-12\)
\(\Leftrightarrow3x=-18\)
\(\Leftrightarrow x=\frac{-18}{3}\)
\(\Leftrightarrow x=-6\)
\(\text{Vậy }x=-6\)