so sánh: 2018^100+2018^99 và 2019^100
Bài toán : So sánh A và B
\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)
\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)
Bài toán : So sánh A và B
\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)
+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)
\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)
+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)
\(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)
.....................................
\(1=1\)
\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
1. So sánh A và B biết : A = \(\frac{2019^{2019}+1}{2019^{2020}+1}\) ; B =\(\frac{2019^{2018}+1}{2019^{2019}+1}\)
2.So sánh M và N biết: M = \(\frac{100^{100}+1}{100^{99}+1}\) ; N= \(\frac{100^{101}+1}{100^{100}+1}\)
Hiện tại mình đang cần gấp giúp mk nha!
1
\(A=\frac{2019^{2019}+1}{2019^{2020}+1}< \frac{2019^{2019}+1+2018}{2019^{2020}+1+2018}=\frac{2019^{2019}+2019}{2019^{2020}+2019}=\frac{2019\left(2019^{2018}+1\right)}{2019\left(2019^{2019}+1\right)}\)
\(=\frac{2019^{2018}+1}{2019^{2019}+1}\)
2
\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)
\(=\frac{100^{100}+1}{100^{99}+1}=N\)
So sánh hai phân số: 2018^100+1/ 2018^90+1 và 2018^99+1/ 2018^89+1
so sánh:
A=\(\frac{100^{2017}+1}{100^{2018}+1}\)với B=\(\frac{100^{2018}+1}{100^{2019}+1}\)
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100\cdot\left[100^{2017}+1\right]}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100\cdot\left[100^{2018}+1\right]}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)
Tự so sánh
\(A=\frac{100^{2017}+1}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)
\(\Rightarrow100A=\frac{100^{2018}+1}{100^{2018}+1}+\frac{99}{100^{2018}+1}\)
\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)(1)
\(B=\frac{100^{2018}+1}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)
\(\Rightarrow100B=\frac{100^{2019}+1}{100^{2019}+1}+\frac{99}{100^{2019}+1}\)
\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)(2)
Từ (1) và (2) suy ra 100A > 100B hay A > B
\(B=\frac{100^{2018}+1}{100^{2019}+1}< \frac{100^{2018}+1+99}{100^{2019}+1+99}=\frac{100^{2018}+100}{100^{2019}+100}=\frac{100\left(100^{2017}+1\right)}{100\left(100^{2018}+1\right)}=\frac{100^{2017}+1}{100^{2018}+1}=A\)
\(\Rightarrow B< A\)
E =\(\frac{2018^{99}-1}{2018^{100}-1}\)so sánh với F =\(\frac{2018^{98}-1}{2018^{99}-1}\)
Ai nhanh tk
Ta có \(E=\frac{2018^{99}-1}{2018^{100}-1}\)
\(\Leftrightarrow2018E=\frac{2018^{100}-2018}{2018^{100}-1}\)
\(\Leftrightarrow2018E=1-\frac{2017}{2018^{100}-1}\) (2)
Lại có \(F=\frac{2018^{98}-1}{2018^{99}-1}\)
\(\Leftrightarrow2018F=\frac{2018^{99}-2018}{2018^{99}-1}\)
\(\Leftrightarrow2018F=1-\frac{2017}{2018^{99}-1}\) (2)
Mà \(2018^{100}>2018^{99}>0\)
\(\Leftrightarrow2018^{100}-1>2018^{99}-1\)
\(\Leftrightarrow\frac{2017}{2018^{100}-1}< \frac{2017}{2018^{99}-1}\)
\(\Leftrightarrow-\frac{2017}{2018^{100}-1}>-\frac{2017}{2018^{99}-1}\)
\(\Leftrightarrow1-\frac{2017}{2018-1}>1-\frac{2017}{2018^{99}-1}\) (3)
Từ (1) ;(2) và (3) <=> 2018E > 2018 F > 0
<=> E > F
Vậy E > F
@@ Học tốt
Chiyuki Fujito
K cần tk
Cho M=(2018^2018+2019^2018)^2019 và N=(2018^2019+2019^2019)^2018. So sánh M và N
So sánh
A = 2018 99 -1 phần 2018100 - 1
B = 201898 -1 phần 201899 - 1
so sánh A=2018^2019 -1/2018^2019+1 và B = 2018^2019/2018^2019+2
Ta có: B = (2018 + 2019)/(2019 + 2020) = (2018 + 2019)/4039 = 2018/4039 + 2019/4039
Ta thấy : 2018/2019 > 2018/4039
2019/2020 > 2019/4039
=> 2018/2019 + 2019/2020 > 2018/4039 > 2019/4039
=> 2018/2019 + 2019/2020 > (2018 + 2019)/(2019 + 2020)
=> A > B
So sánh:
\(C=\dfrac{2019-2018}{2018+2019}\) và \(D=\dfrac{2019^2-2018^2}{2019^2+2018^2}\)
Ta có: \(C=\dfrac{2019-2018}{2019+2018}\)
\(\Leftrightarrow C=\dfrac{\left(2019-2018\right)\left(2019+2018\right)}{\left(2019+2018\right)^2}\)
\(\Leftrightarrow C=\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}\)
Ta có: \(\left(2019+2018\right)^2=2019^2+2018^2+2\cdot2019\cdot2018\)
\(2019^2+2018^2=2019^2+2018^2+0\)
Do đó: \(\left(2019+2018\right)^2>2019^2+2018^2\)
\(\Leftrightarrow\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}< \dfrac{2019^2-2018^2}{2019^2+2018^2}\)
\(\Leftrightarrow C< D\)