Lời giải:
$\frac{3}{4}-(x+1\frac{1}{2})=(-1)^{2024}=1$
$x+\frac{3}{2}=\frac{3}{4}-1=\frac{-1}{4}$
$x=\frac{-1}{4}-\frac{3}{2}=\frac{-7}{4}$
Tìm x,y biết
a,\(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=994-15:3+1^{2025}\)
b,\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
c,\(2024^{|x-1|+y^2-1}\cdot3^{2024}=9^{1012}\)
tìm x,y biết
a,\(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
b,\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
c,\(2024^{\left|x-1\right|=y^2-1}\cdot3^{2024}=9^{1012}\)
1/2 +1/3+1/4+...+ 1/2023+ 1/2024
cho 2 số x, y thỏa mãn (y-1) ^2024+|x+y-1|=0 tính giá trị của biểu thức A=x^2024+y^2024
(2x-1)/2020+(2x-3)/2024= (1-2 x)/1008
tìm x biết |2x-1|+(2/3-x)^2024=0
\(\dfrac{x-5}{3}=\dfrac{-12}{5-x}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)
(-1)-1/2-1/4-/1/6-.....-1/2024
D = 1+ 1/2+1/4+1/8+1/16+...+1/2024
