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\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
A = 3004.3004 và B = 3000.3008
Ta có:
A = 3004.3004
A = 3004.(3000 + 4)
A = 3004.3000 + 3004.4
A = 3004.3000 + 12016
B = 3000.3008
B = 3000.(3004 + 4)
B = 3000.3004 + 3000.4
B = 3000.3004 + 12000
Vì 3004.3000 = 3000.3004 ; 12016 > 12000 nên A > B
a < b vì 20169 < 201710
câu này rất dễ,chỉ càn lí luận thôi
Bài làm :
\(148:\left(x+2\right)=37\)
\(x+2=148:37\)
\(x+2=4\)
\(x=4-2\)
\(x=2\)
Học tốt
1:
a: \(1\dfrac{2}{3}-\dfrac{7}{12}x=\dfrac{-5}{24}\cdot\dfrac{8}{-15}\)
=>\(\dfrac{5}{3}-\dfrac{7}{12}x=\dfrac{5}{15}\cdot\dfrac{8}{24}=\dfrac{1}{9}\)
=>\(\dfrac{7}{12}x=\dfrac{5}{3}-\dfrac{1}{9}=\dfrac{14}{9}\)
=>\(x=\dfrac{14}{9}:\dfrac{7}{12}=\dfrac{14}{9}\cdot\dfrac{12}{7}=2\cdot\dfrac{4}{3}=\dfrac{8}{3}\)
b: \(\dfrac{25}{-30}=\dfrac{15}{x}\)
=>\(\dfrac{15}{x}=\dfrac{5}{-6}\)
=>\(x=\dfrac{15\cdot\left(-6\right)}{5}=-3\cdot6=-18\)
2:
\(\left(\dfrac{1}{32}\right)^7=\left[\left(\dfrac{1}{2}\right)^5\right]^7=\left(\dfrac{1}{2}\right)^{35}\)
\(\left(\dfrac{1}{16}\right)^9=\left[\left(\dfrac{1}{2}\right)^4\right]^9=\left(\dfrac{1}{2}\right)^{36}\)
mà 35<36 và 1/2<1
nên \(\left(\dfrac{1}{2}\right)^{35}>\left(\dfrac{1}{2}\right)^{36}\)
=>\(\left(\dfrac{1}{32}\right)^7>\left(\dfrac{1}{16}\right)^9\)