so sanh a va b
A = 2017^ 10 x 2017 ^ 9
B= 2018^ 10
Những câu hỏi liên quan
so sanh
A=2016/2017+2017/2018 va B=2016+2017/2017+2018
So sánh \(A=\dfrac{2016}{2017}+\dfrac{2017}{2018}\) và \(B=\dfrac{2016+2017}{2017+2018}\)
Có 2 cách:
C1 :Rảnh thì bấm máy tính luôn rồi so sánh (nhưng cách này tỉ lệ sai khá cao nếu bất cẩn ghi nhầm số):
\(A=\dfrac{2016}{2017}+\dfrac{2017}{2018}\) \(=1,999008674\approx2\)
\(B=\dfrac{2016+2017}{2017+2018}\) \(=0,9995043371\approx1\)
Do 2 > 1 nên :
\(\Rightarrow A>B\).
C2:
Ta có:
\(\dfrac{2016}{2017}>\dfrac{2016}{2018}\Rightarrow A>\dfrac{2016}{2018}+\dfrac{2017}{2018}\Rightarrow A>\dfrac{2016+2017}{2017}\)
\(B=\dfrac{2016+2017}{2017+2018}=\dfrac{2016+2017}{4035}\)
Vì \(\dfrac{2016+2017}{2018}>\dfrac{2016+2017}{4035}\)
\(\Rightarrow A>B\).
_ Học tốt :))_
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so sanh:
a. 2^70 va 3^51
b. 2015/2017 va 2017/2018
351>350=925>825=275>270
Vì 2017<2018 nên\(\frac{1}{2017}\)>\(\frac{1}{2018}\)
⇒\(\frac{2}{2017}\)>\(\frac{1}{2018}\)
⇒\(\frac{2015}{2017}\)=1-\(\frac{2}{2017}\)<1-\(\frac{1}{2018}\)=\(\frac{2017}{2018}\)
Vậy, \(\frac{2015}{2017}\)< \(\frac{2017}{2018}\)
so sanh a/b va a+2017/b+2018
So sanh A=2018/-2017 va B=-2019/2019
2018/-2017<-1=-2019/2019
suy ra 2018/-2017<-2019/2019
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Giải
Ta có: \(\frac{2018}{-2017}=-1+\frac{-1}{2017}\)
\(\Rightarrow\frac{2018}{-2017}>-1\)
Mà \(\frac{-2019}{2019}=1\) nên \(\frac{2018}{-2017}>\frac{-2019}{2019}\)
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so sanh A B biet A=\(\frac{10^{2018}+1}{10^{2017}+1}\);B=\(\frac{10^{2007}+1}{10^{2008+1}}\)
Ta có: \(10^{2018}>10^{2017}\Rightarrow10^{2018}+1>10^{2017}+1\Rightarrow A=\frac{10^{2018}+1}{10^{2017}+1}>1\) (1)
\(10^{2007}< 10^{2008}\Rightarrow10^{2007}+1< 10^{2008}+1\Rightarrow B=\frac{10^{2007}+1}{10^{2008}+1}< 1\) (2)
Từ (1) và (2) => A > B
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So sanh a va b ma khong tinh gia tri cua a va b
A=2016×2018 ; B = 2017× 2017
Ai lam đung minh se tich cho😙😙😙😙😙😙
Ta có :
\(A=2016.2018\)
\(\Rightarrow A=2016\left(2017+1\right)\)
\(\Rightarrow A=2016.2017+2016\)
Ta lại có :
\(B=2017.2017\)
\(\Rightarrow B=2017.\left(2016+1\right)\)
\(\Rightarrow B=2017.2016+2017\)
Ta thấy: \(2017>2016\)
\(\Rightarrow2017.2016+2017>2017.2016+2016\)
\(\Rightarrow B>A\)
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Hãy so sánh:
A=10^2016+2018/10^2017+2018
B=10^2017+2018/10^2018+2018
nhanh lên các bn mik cần gấp
Không dùng máy tính hãy so sánh A=10^2016+2018/10^2017+2018 và B=10^2017+2018/10^2018+2018
Ta có: \(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)\(\Rightarrow10A=\frac{10^{2017}+2018.10}{10^{2017}+2018}=\frac{10^{2017}+2018+2018.9}{10^{2017}+2018}=1+\frac{2018.9}{10^{2017}+2018}\)
Tương tự ta có: \(10B=1+\frac{2018.9}{10^{2018}+2018}\)
Vì \(2017< 2018\)\(\Rightarrow10^{2017}< 10^{2018}\)\(\Rightarrow10^{2017}+2018< 10^{2018}+2018\)
\(\Rightarrow\frac{2018.9}{10^{2017}+2018}>\frac{2018.9}{10^{2018}+2018}\)\(\Rightarrow1+\frac{2018.9}{10^{2017}+2018}>1+\frac{2018.9}{10^{2018}+2018}\)
hay \(10A>10B\)\(\Rightarrow A>B\)
Vậy \(A>B\)
Ta có : \(A=\frac{10^{2016}+2018}{10^{2017}+2018}\)
\(\Rightarrow10A=\frac{10^{2017}+20180}{10^{2017}+2018}=\frac{10^{2017}+2018+18162}{10^{2017}+2018}=1+\frac{18162}{10^{2017}+2018}\)
Ta có : \(B=\frac{10^{2017}+2018}{10^{2018}+2018}\)
\(\Rightarrow\frac{10^{2018}+20180}{10^{2018}+2018}=\frac{10^{2018}+2018+18162}{10^{2018}+2018}=1+\frac{18162}{10^{2018}+2018}\)
Vì \(10^{2017}+2018< 10^{2018}+2018\) nên \(\frac{18162}{10^{2017}+2018}>\frac{18162}{10^{2018}+2018}\)
\(\Rightarrow1+\frac{18162}{10^{2017}+2018}>1+\frac{18162}{10^{2017}+2018}\Rightarrow10A>10B\Rightarrow A>B\)
Vậy A > B
Làm khác bạn kia 1 xíu à
so sanh 2017^2018 va 2018^2019
ta thấy\(1-\frac{2017}{2018}=\frac{1}{2018}>\frac{1}{2019}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{2017}{2018}< \frac{2018}{2019}\)
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