Môn Toán Lớp 6: Tìm x biết :
1) 2x – 1 : 4 = 16
2) (3x – 1 )^10 = ( 3x – 1 )^20
3) ( 7x – 1)^3 = 2^5 . 5^2 + 200
4) (2x + 1 )^3 = 125
5) x ( 5 – x )^3 = ( 5-x )^3
Môn Toán Lớp 6: Tìm x biết : 1) 2x – 1 : 4 = 16 2) (3x – 1 )^10 = ( 3x – 1 )^20 3) ( 7x – 1)^3 = 2^5 . 5^2 + 200 4) (2x + 1 )^3 = 125 5) x (
2x -1 = 4
2x = 5
x = 2,5
2) (3x – 1 )^(10) = ( 3x – 1 )^(20)
⇔ (3x – 1 )^10 [ (3x – 1 )^10 – 1] = 0
3) ( 7x – 1)^3 = 2^5 × 5^2 + 200
Biến đổi vế trái của phương trình
$\text{(7x-1)^3 = 7x-1}$
(7x-1) = 2^3 × 5^3
7x-1 = 1000
7x = 1001
x = 143
4) (2x + 1 )^3 = 125
(2x + 1 )^3 = 5^3
2x +1 = 5
2x = 4
x = 2
5) x ( 5 – x )^3 = ( 5-x )^3
⇒$\left[\begin{matrix} x = 1\\ 5-x=0\end{matrix}\right.$
⇒$\left[\begin{matrix} x=1\\ x=5\end{matrix}\right.$
2x-1 : 4 = 16
=>2x – 1/4 = 16
=>2x=16+1/4
=>2x =65/4
=>x = 65/4 : 2
=>x=65/8
Vậy x=65/8
2,
(3x-1)^{10} = (3x-1)^{20}
<=>(3x-1)^{20} – (3x-1)^{10}=0
<=>(3x-1)^{10}[(3x-1)^{10}-1]=0
<=>\(\left[ \begin{array}{l}(3x-1)^{10}=0\\(3x-1)^{10} – 1 = 0\end{array} \right.\) <=>\(\left[ \begin{array}{l}3x-1=0\\\left[ \begin{array}{l}3x-1=1\\3x-1=-1\end{array} \right. \end{array} \right.\) <=>\(\left[ \begin{array}{l}x=\dfrac{1}{3}\\\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=0\end{array} \right. \end{array} \right.\)
Vậy x \in {1/3;2/3;0}
3,
(7x-1)^{3} = 2^{5}.5^{2}+200
=>(7x-1)^{3} = 800 + 200
=>(7x-1)^{3} = 1000
=>(7x-1)^{3} = 10^{3}
=>7x-1=10
=>7x=11
=>x=11/7
Vậy x=11/7
4,
(2x+1)^{3} = 125
=>(2x+1)^{3} = (5)^{3}
=>2x+1=5
=>2x=4
=>x=2
Vậy x = 2
5,
x(5-x)^{3} = (5-x)^{3}
<=>\(\left[ \begin{array}{l}x=1\\5-x=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=1\\x=5\end{array} \right.\)
Vậy x \in {1;5}