Giúp mình với
So sánh
a, \(\frac{-21}{25}\)VÀ \(\frac{-123}{167}\)
b, \(\frac{1999}{2000}\) VÀ \(\frac{199}{200}\)
Giup mình so sánh bài này với
\(\frac{1999}{2000}\)VÀ \(\frac{199}{200}\)
Viết cho mình cách làm nhé , cô giáo mình bắt viết cả cách làm
\(\frac{1999}{2000}=1-\frac{1}{2000}\) (1)
\(\frac{199}{200}=1-\frac{1}{200}\) (2)
\(\frac{1}{200}>\frac{1}{2000}\) (3)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\frac{1999}{2000}>\frac{199}{200}\)
\(S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{\frac{1}{1999}+\frac{2}{1998}+...+\frac{198}{2}+\frac{199}{1}}\) giải giúp mình với nhé
\(M=1+\frac{1}{199}+1+\frac{2}{198}+1+....+\frac{198}{2}+1=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}\)
\(=200.\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)=200 T
\(S=\frac{T}{200T}=\frac{1}{200}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+...+\frac{198}{2}+\frac{199}{1}}\)
Giúp mình với bạn nào trả lời nhanh và chính xác nhất mình sẽ tích cho và hạn nộp trước 11h nhé
Để chiều mình làm cho
làm luôn đi bạn mình đang cần vội
MS=
\(\frac{1}{199}+\frac{2}{198}+...+\frac{198}{2}+199\)
=\(\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)
=\(\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}+\frac{200}{200}\)
=\(200.\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}+\frac{1}{200}\right)\)
= 200. TS
\(\Rightarrow\)Phân số đã cho = \(\frac{1}{200}\)
Chú ý: MS là mẫu, TS là tử
Không tính , hãy so sánh :
\(A=\frac{199}{200}+\frac{200}{201}+\frac{201}{202}\)
\(B=\frac{199+200+201}{200+201+202}\)
\(\frac{199}{200}>\frac{199}{200+201+202}\)
\(\frac{200}{201}>\frac{200}{200+201+202}\)
\(\frac{201}{202}>\frac{201}{200+201+202}\)
=>\(A>B\)
Do \(\frac{199}{200}\)> \(\frac{199}{200+201+202}\), \(\frac{200}{201}\)>\(\frac{200}{200+201+202}\),\(\frac{201}{202}\)>\(\frac{201}{200+201+202}\)nên A>B
Chứng tỏ rằng :
1\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}>\frac{25}{12}\)
Ai làm đúng và nhanh nhất mình tick cho nhé ! (:D)
thực ra nó rất là dễ. giờ mình mới phát hiện ra chứ bữa trước mình làm cách dài lắm
Ta có :
\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)\)
\(=\frac{25}{12}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)>\frac{25}{12}\)( đpcm )
Không tính , hãy so sánh :
\(A=\) \(\frac{199}{200}+\frac{200}{201}+\frac{201}{202}\)
\(B=\frac{199+200+201}{200+201+202}\)
\(A=\frac{199}{200}+\frac{200}{201}+\frac{201}{202}< \frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}\)
A \(< \frac{199+200+201}{200+201+202}=B\)
\(A< B\)
Ta có: \(A=\frac{199}{200}+\frac{200}{201}+\frac{201}{202}< \frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}< \)
\(< \frac{199+200+201}{200+201+202}\)
Vậy A < B
ỦNG HỘ TỚ NHA
\(B=\frac{199+200+201}{200+201+202}\)
\(B=\frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}< \frac{199}{200}+\frac{200}{201}+\frac{201}{202}=A\)
Vậy B < A ( bài này chủ yếu so sánh hai phân số cùng từ, phân số nào có mẫu lớn hơn thì phân số đó nhỏ hơn )
1. Tính nhanh:
a.\(\frac{17}{13}\times\frac{7}{15}-\frac{5}{12}\times\frac{21}{39}+\frac{49}{91}\times\frac{8}{15}\)
b.\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
2. So sánh:
a. 3200và2300
b. 7150và3775
c.\(\frac{201201}{202202}\)và\(\frac{201201201}{202202202}\)
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
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(x-20).\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)=\(\frac{1}{2000}\)
Giúp mk với
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)
\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)
So sánh: C=\frac{1999^2000+1/1999^1999+1} và D=\frac{1999^1999+1/1999^1998+1}