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,\(2024^{\left|x-1\right|=y^2-1}\cdot3^{2024}=9^{1012}\)
a,4.|3x-1|=|6x-2|+|-1,5|
b,2024.|2x-1|=2025.|1-2x|-|-2|
c,|2x+1|+|3x-1|=0
SO SÁNH 45 VỚI S
\(S=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}+...+\frac{1}{\sqrt{2025}-\sqrt{2024}}\)
Tìm số dư trong phép chia\(\left(2023^{2024}+2024^{2025}+2025^{2026}\right)^{10}\) chia cho 111
cho các số thực x,y,z thỏa mãn \(\left(x-y +z\right)^2\)+\(\sqrt{y^4}\)+\(\left|1-z^3\right|\) \(\le\) 0
Chứng minh rằng \(x^{2023}\)+\(y^{2024}\)+\(z^{2025}\)=0
Tìm số dư trong phép chia (2023\(\left(2023^{2024}+2024^{2025}+2025^{2026}\right)^{10}\)chia cho 111
So sánh 45 với S, biết:
S= \(\frac{1}{\sqrt{1}}\)+ \(\frac{1}{\sqrt{2}-\sqrt{1}}\)+ \(\frac{1}{\sqrt{3}-\sqrt{2}}\)+ \(\frac{1}{\sqrt{4}-\sqrt{3}}\)+....+ \(\frac{1}{\sqrt{2025}-\sqrt{2024}}\)
So sánh:
a)\(\frac{2015}{2016}v\text{à}\frac{2035}{2034}\) b) \(\frac{-2025}{2024}v\text{à}\frac{-2026}{2027}\)
3/4-(x+1(1/2))=(-1)^2024