SO SÁNH CÁC SỐ HỮU TỈ SAU:
-2016/2017 VÀ 1/10
99/-100 VÀ -102/101
GIÚP MÌNH VỚI
so sánh 2 số hữu tỉ sau -99/100 vs -102/101
ban co the so sanh bang cach quy dong mau hoac tu vi cac cach khac ko giai dc.
So sánh các PS sau :
a) 99/100 và 100/99
b) 99/100 và 100/101
c) 24/50 và 50/97
d)2015/2016 và 2017/2018
e) 5/16 và 501/1601
f) 249/500 và 500/997
nhanh lên mai mình phải nộp bài rồi
Trả lời :
a)\(\frac{99}{100}< 1\)và \(\frac{100}{99}>1\)nên \(\frac{99}{100}< \frac{100}{99}\)
~ Hok tốt ~
b,....đề....
ta có :1-\(\frac{99}{100}\)=\(\frac{1}{100}\)
1-\(\frac{100}{101}\)= \(\frac{1}{101}\)
mà \(\frac{1}{100}\) > \(\frac{1}{101}\)
=> \(\frac{99}{100}\) > \(\frac{100}{101}\)
SO SÁNH :52017 và 251008
Cho A=10101-1 /10102-1;B=10100+1/10101+1.SO SÁNH A và B
ta có :
\(25^{1008}=\left(5^2\right)^{1008}=5^{2.1008}=5^{2016}\)
mà \(5^{2017}>5^{2016}\)
\(\Rightarrow\)\(5^{2017}>\left(5^2\right)^{1008}\)
\(\Rightarrow\)\(5^{2017}>25^{1008}\)
có \(5^{2017}=\left(5^2\right)^{1008}\times5\)\(=25^{1008}\times5\)
mà \(=25^{1008}\times5\)> \(25^{1008}\)
nên \(5^{2017}>25^{1008}\)
Ta có:
\(5^{2017}>5^{2016}=\text{[}5^2\text{]}^{1008}=25^{1008}\)
Suy ra: 52017 > 251008
Ta có:
\(1-A=1-\frac{10^{101}-1}{10^{102}-1}=\frac{10^{102}-1-\text{[}10^{101}-1\text{]}}{10^{102}-1}=\frac{10^{102}-1-10^{101}+1}{10^{102}-1}\)\(=\frac{10^{102}-10^{101}}{10^{102}-1}=\frac{10^{101}\left[10-1\right]}{10^{101}\text{[}10-\frac{1}{10^{101}}\text{]}}=\frac{10-1}{10-\frac{1}{10^{101}}}=\frac{9}{10-\frac{1}{10^{101}}}\)
\(1-B=1-\frac{10^{100}+1}{10^{101}+1}=\frac{10^{101}+1-\left[10^{100}+1\right]}{10^{101}+1}=\frac{10^{101}+1-10^{100}-1}{10^{100}+1}\)
\(=\frac{10^{101}-10^{100}}{10^{101}+1}=\frac{10^{100}\left[10-1\right]}{10^{100}\text{[}10+\frac{1}{10^{100}}\text{]}}=\frac{10-1}{10+\frac{1}{10^{100}}}=\frac{9}{10+\frac{1}{10^{100}}}\)
Vì \(\frac{9}{10-\frac{1}{10^{101}}}>\frac{9}{10+\frac{1}{10^{100}}}\Rightarrow A< B\)
So sánh A và B, biết:
A=(10100+1) : (10101+1)
B=(10101+1) : (10102+1)
Các bạn giải giúp mình, cảm ơn nhiều!
B=(10101+1):(10102+1)<(10101+1+9):(10102 +1+9)=(10101+10):(10102+10)=[10.(10100+1]:[10.(10101+)]
=(10100+1):(10101+1)=A
=>A>B
So sánh ps : 2017^99 + 1/2017^100 + 1 và 2017^100 + 1/2017^101 + 1
Ta có: \(A=\frac{2017^{99}+1}{2017^{100}+1}\Rightarrow2017A=\frac{2017^{100}+2017}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\Rightarrow2017B=\frac{2017^{101}+2017}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
\(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
\(\Rightarrow2017A>2017B\Rightarrow A>B\)
Vậy...
Đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\)nên \(2017A=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(B=\frac{2017^{100}+1}{2017^{101}+1}\)nên \(2017B=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
Vì \(1=1;\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\)
Hay \(2017A>2017B\)nên \(A>B\)
Vây \(\frac{2017^{99}+1}{2017^{1001}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)
đặt \(A=\frac{2017^{99}+1}{2017^{100}+1}\); \(B=\frac{2017^{100}+1}{2017^{101}+1}\)
Ta có : \(2017A=\frac{2017.\left(2017^{99}+1\right)}{2017^{100}+1}=\frac{2017^{100}+2017}{2017^{100}+1}=\frac{2017^{100}+1+2016}{2017^{100}+1}=1+\frac{2016}{2017^{100}+1}\)
\(2017B=\frac{2017.\left(2017^{100}+1\right)}{2017^{101}+1}=\frac{2017^{101}+2017}{2017^{101}+1}=\frac{2017^{101}+1+2016}{2017^{101}+1}=1+\frac{2016}{2017^{101}+1}\)
Vì \(\frac{2016}{2017^{100}+1}>\frac{2016}{2017^{101}+1}\Rightarrow1+\frac{2016}{2017^{100}+1}>1+\frac{2016}{2017^{101}+1}\Leftrightarrow10A>10B\Rightarrow A>B\)
So Sánh \(\frac{2017^{99}+1}{2017^{100}+1}\)và \(\frac{2017^{100}+1}{2017^{101}+1}\)
Giải giùm mình với
vì 2017100 + 1 < 2017101 + 1
\(\Rightarrow\frac{2017^{100}+1}{2017^{101}+1}< \frac{2017^{100}+1+2016}{2017^{101}+1+2016}=\frac{2017^{100}+2017}{2017^{101}+2017}=\frac{2017.\left(2017^{99+1}\right)}{2017.\left(2017^{100}+1\right)}=\frac{2017^{99}+1}{2017^{100}+1}\)
Vậy \(\frac{2017^{99}+1}{2017^{100}+1}>\frac{2017^{100}+1}{2017^{101}+1}\)
so sánh 2 phân số cùng mẫu thì ta xét tử
đừng nói không làm được chứ
so sánh 2017^99+1/2017^100+1 và 2017^100+1/2007^101+1
Bài 1 : Tính nhanh
a) 6/15 + 6/35 + 6/63 + 6/99 + 6/143
b) 3/24 + 3/48 + 3/80 + 3/120 + 3/168
Bài 2 : So sánh các phân số sau
a) 2/3 và 5/6 b) 1/4 và 151515/101010 c) 2017/2016 và 2017/2018 d) 2014/2015 và 2015/2016
Bài 3 : So sánh
B = 1/51 + 1/52 + ..... + 1/99 + 1/100 và 1/2
Giải bài giải đầy đủ giúp mình nhé
1.
a) \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)
\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{6}{2}.\frac{10}{39}\)
\(=\frac{10}{13}\)
b) \(\frac{3}{24}+\frac{3}{48}+\frac{3}{80}+\frac{3}{120}+\frac{3}{168}\)
\(=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)
\(=\frac{3}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\frac{5}{28}\)
\(=\frac{15}{56}\)
\(a.\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{11.13}\)
\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=3.\frac{10}{39}\)
\(=\frac{10}{13}\)
\(a.\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)
\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)
\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=3.\frac{10}{39}\)
\(=\frac{10}{13}\)
So sánh các phân số 100/101+101/102 và 100+101/101+102
Ta có: 100+101/101+102
= 100/101+102 + 101/101+102
Vì 100/101>100/101+102
101/102 > 101/101+102
=>100/101+101/102 > 100+101/101+102