Tính :
\(\left(\frac{12}{199}+\frac{23}{200}\right)\)\(x\)\(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Những câu hỏi liên quan
\(a.\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0\)
\(=0\)
Kết quả = 0 nhé, nhớ ủng hộ mh, mh đang âm diểm
~ HOK TỐT ~
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\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0\)
\(=0\)
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\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).0\)
\(=0\)
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Tính :
\(\left(\frac{12}{199}+\frac{23}{200}\right)\)\(x\)\(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Các bạn giúp mình nha !!!
Không quy đồng , hãy tính hợp lí tổng sau
C = \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\) . \(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Ta thay 1/2 - 1/3 - 1/6 = 0 ma khi 1 thua so nao nhan voi 0 thi h se la 0
Vay C= 0
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\(C=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(C=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).0=0\)
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Tính nhanh:
a,\(\frac{7}{13}\cdot\frac{7}{15}-\frac{5}{12}\cdot\frac{21}{39}+\frac{49}{91}\cdot\frac{8}{15}\)
b,\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
1. Tính nhanh
a, \(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{91}.\frac{8}{15}\)
b, \(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
c,\(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
1.Tính:
\(a.\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{91}.\frac{8}{15}\)
\(b.\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
2.So sánh
\(a.3^{200}và2^{300}\)
\(b.71^{50}và37^{75}\)
Tính bằng cách hợp lí:
a) A=\(\left(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+....+\frac{1}{101.400}\right):\left(\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+...+\frac{1}{299.400}\right)\)
b) B=\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{200}\right):\left(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....\frac{198}{2}+\frac{199}{1}\right)\)
S= 1/199 + 2/198 + ... + 198/2 + 199/1
S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2 + 1) +1
S= 200/200 + 200/199 + 200/198 + ... + 200/2
S= 200.(1/200 + 1/199 + ... + 1/2)
Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)
B=1 : 200 = 1/200
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1. Tính nhanh:a.frac{17}{13}timesfrac{7}{15}-frac{5}{12}timesfrac{21}{39}+frac{49}{91}timesfrac{8}{15}b.left(frac{12}{199}+frac{23}{200}-frac{34}{201}right)timesleft(frac{1}{2}-frac{1}{3}-frac{1}{6}right)2. So sánh:a. 3200và2300b. 7150và3775c.frac{201201}{202202}vàfrac{201201201}{202202202}
Đọc tiếp
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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\(ChoA=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{200}-1\right)HãysosánhA\:với\frac{-1}{199}\)
A = \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{200}-1\right)\)
= \(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-199}{200}\)
= \(\frac{-1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{199}{200}\)
= \(\frac{-1}{200}\)> \(\frac{-1}{199}\)( vì 1/200 < 1/999 => - 1 / 200 > -1/199 )
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