tìm x biết: |2x-1|+1=4
Tìm x biết:
a)|2x+1|=|x+4|
b) |2x-1|=x+4
a) \(\left|2x+1\right|=\left|x+4\right|\Rightarrow\left[{}\begin{matrix}2x+1=x+4\\2x+1=-x-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b) \(\left|2x-1\right|=x+4\Rightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Tìm x biết √(2x - 1) - √(x +1 ) = 2x - 4
Tìm x biết √(2x - 1) - √(x +1 ) = 2x -4
tìm x biết b)(x-1)(3-2x)+(2x-1)(x+3)=4
=>3x-2x^2-3+2x+2x^2+6x-x-3=4
=>10x=10
=>x=1
Tìm x biết
1, | 2x + 4 | +| 4 - x | = 11
2, | x | + | x - 1 | + | 2x - 4 | = 3
1) =>2x+4+4+x=11
=>2x+4+4+x-11=0
=>3x-3=0
=>3x=3
=> x=1
Vậy x thuộc {1}
2)=>x+x+1+2x+4=3
=>x+x+1+2x+4-3=0
=>4x+2=0
=>4x=-2
=>x=-2/4
=>x=-1/2
Vậy x thuộc {-1/2}
Tìm x biết: (2x-1)^5 = (2x-1)^4
\(\Leftrightarrow\left(2x-1\right)^4\left(2x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
tìm x, biết
(2x+1)\(^4\)=(2x+1)\(^6\)
0=(2x+1)2
4x2 + 4x + 1 = 0
4x2 = 0 hay 4x + 1 = 0
x = 2 hay x= \(-\dfrac{1}{4}\)
(2x+1)=(2x+1)
=> (2x+1)^4 - (2x+1)^6 = 0
=> (2x+1)^4 * [1 - (2x+1)^2] = 0
=> \(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left[1-\left(2x+1\right)^2\right]=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x=0\\2x=-2\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)Vậy x\(\in\){0;-1;\(\dfrac{1}{2}\)}
Tìm x biết: giúp với mn ơi
a) (x - 2)(x² + 2x+ 4) + x(x + 3)(3 - x) = 1
b) (2x + 1)³ - (2x - 1)³ - 6(2x - 1)² = 5
\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)
a) \(\Leftrightarrow x^3-8-x^3+9x=1\)
\(\Leftrightarrow9x=9\Leftrightarrow x=1\)
b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)
\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)
Tìm số tự nhiên x, biết:
a) ( x + 1 ) 4 = ( 2 x ) 4 ;
b) ( 2 x - 1 ) 5 = x 5
a) Ta có: ( x + 1 ) 4 = ( 2 x ) 4 nên x +1 = 2x. Do đó x = 1.
b) Ta có: ( 2 x - l ) 5 = x 5 nên 2x - l = x. Do đó x = l.
Tìm X biết: ( 9 x 2x - 1)^2 =(1-2x)^4