tìm x biết: ( 2x + 1 ) 4 = ( 2x + 1 )6
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 x biết:(2x+1)^4=(2x+1)^6
Bài 2 Tìm x biết 1, (2x-2).(3x+1)-(3x-2).(2x-3)=5 2,(1-3x).(3x-5)-(2x-4)(2-3x)=x-6 3,(2x-1).(4x^2+2x+1)-(2x+1)(4x^2-2x+1)=5x+6 Giúp tớ với
Bài 1: Tìm số hữu tỉ x biết:
a, ( 2x - 1 )4 = 81 b, ( x - 1 )5 = -32
c, ( 2x - 1 )6 = ( 2x - 1 )8
Bài 2: Tìm các số tự nhiên x, y biết rằng:
a, 2x + 1 . 3y = 12x. b, 10x : 5y = 20y
c, 2x = 4y - 1 và 27y = 3x + 8
Bài 2:
a: Ta có: \(2^{x+1}\cdot3^y=12^x\)
\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Tìm x biết:
a; 3x.(2x+3)-(2x+5x).(3x-2)=8
b;4.(x-1)-3.(x^2-5)_x^2=(x-3)-(x+4)
c; 2.(3x-1).(2x+5)-6.(2x-1).(x+2)=-6
d; 3.(2x-1).(3x-1)-(2x-3.(9x-1)-3=3
Tìm x biết:
a,2x(x+1)-3-2x=5
b,2x(3x+1)+(4-2x)=7\
c,(x-3)^3-(x-3)(x^2+3x+9)+6(x-1)^2=6
a)\(2x\left(x+1\right)-3-2x=5\)
\(\Leftrightarrow2x^2+2x-3-2x=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4=\left(-2\right)^2=2^2\)
\(\Rightarrow x=2;-2\)
b)\(2x\left(3x+1\right)+\left(4-2x\right)=7\)
\(\Leftrightarrow6x^2+2x+4-2x=7\)
\(\Leftrightarrow6x^2+4=7\)
\(\Leftrightarrow6x^2=3\)
\(\Leftrightarrow x^2=\frac{1}{2}=-\sqrt{\frac{1}{2}}=\sqrt{\frac{1}{2}}\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x-1\right)^2=6\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2-2x+1\right)=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow3x^2+15x=0\)
\(\Leftrightarrow3x\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Tìm x biết:
a,2x(x+1)-3-2x=5
b.2x(3x+1)+(4-2x)=7
c,(x-3)^3-(x-3)(x^2+3x+9)+6(x-1)^2=6
5: Tìm x biết a) x/3 =4/12 b) x-1/ x-2=3/5 c) 2x :6=1/4 d) x² +x/2x²+1=1/2
a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
Tìm x biết : (2x + 1)4 = (2x + 1)6
Vì 2 số có cơ số bằng nhau và số mũ khác nhau nhưng giá trị bằng nhau nên cơ số chỉ có thể là 0 hoặc 1
Nếu giá trị = 0 => x = (0 - 1) : 2 = -0,5
Nếu giá trị = 1 => x = (1 - 1) : 2 = 0
Kết luận : x = -0,5 hoặc = 0