tim x (x+3)(2x-1)-(x-4)(2x+1)=10
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a) Đặt x^2+2x+2=t
\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)
\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)
Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
\(\left|x-\dfrac{1}{2}\right|\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\\\left|2x-\dfrac{3}{4}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow2x-\dfrac{3}{4}\ge0\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1\Rightarrow x=\dfrac{3}{2}\\x-\dfrac{1}{2}=-1\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
\(2x-\dfrac{3}{4}\ge0\Rightarrow2x\ge\dfrac{3}{4}\Rightarrow x\ge\dfrac{3}{2}\)
Vậy xảy ra khi:
\(x=\dfrac{3}{2}\)
12x/6 + 4x/6 - 3x/6 = 3/8 - 1/4 - 1/6
13x/6 = (9 - 6 - 4)/24 = -1/24
--> x = -1/52
|x-3|+1=x
|x+3| = x-1
Th1: x+3 = x-1
0x = 4
x thuộc rỗng
Th2: -(x+3) = x-1
-x -3 = x-1
2x = -2
x= -1
Vậy x=-1
|x-3|=2x+4
Th1: x-3 = 2x +4
x = -7
Th2: -(x-3) = 2x+4
-x +3 = 2x +4
-3x = 1
x= -1/3
Vậy x= -7; x= -1/3
|3-2x|=x+1
Th1: 3 - 2x = x+1
3x = 2
x= 2/3
Th2: - (3-2x) = x+1
- 3 +2x = x+1
x= 4
Vậy x= 2/3 và x= 4
|x-1|+3x=1
|x+1| = 1- 3x
Th1: x+1 = 1- 3x
4x = 0
x=0
Th2: -(x+1) = 1- 3x
-x -1 = 1- 3x
2x = 2
x=1
Vậy x=0 và x=1
\(\left(x+3\right)\left(2x-1\right)-\left(x-4\right)\left(2x+1\right)=10\)
\(\Leftrightarrow2x^2-x+6x-3-\left(2x^2+x-8x-4\right)=10\)
\(\Leftrightarrow2x^2-x+6x-3-2x^2-x+8x+4=10\)
\(\Leftrightarrow12x+1=10\)
\(\Leftrightarrow12x=9\)
\(\Leftrightarrow x=\frac{9}{12}\)
\(\Rightarrow x=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)