2.3/7+(2/9-1 3/7)-5/3:1/9
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2.3/7+(2/9-10/7)-5/3:1/9
\(\frac{2,3}{7}+\left(\frac{2}{9}-\frac{10}{7}\right)-\frac{5}{3}:\frac{1}{9}\)
\(\frac{2,3}{7}-\frac{76}{63}-15\)
\(-\frac{55,3}{63}-15\)
\(-15\frac{55,3}{63}\)
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Tính hợp Lý:
2.3/7+2/9-10/7-5/3:1/9
Tìm x
a) x-2/5+7/12=1/2.3/4.-9/16
b) 4/9.x-1/2=(-5)/9
c) -2/3.x+5/8=7/12
d) 2/5.x:7/3.1/8=29/15
giúp em với mai em nộp rồi
-1/9.-3/5+5/-6.-3/5-7/2.3/5
- \(\dfrac{1}{9}\) \(\times\) ( - \(\dfrac{3}{5}\)) + \(\dfrac{5}{-6}\) \(\times\) ( \(-\dfrac{3}{5}\)) - \(\dfrac{7}{2}\) \(\times\) \(\dfrac{3}{5}\)
= \(\dfrac{1}{9}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{5}{6}\) \(\times\) \(\dfrac{3}{5}\) - \(\dfrac{7}{2}\) \(\times\) \(\dfrac{3}{5}\)
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{1}{9}\) + \(\dfrac{5}{6}\) - \(\dfrac{7}{2}\))
= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{2}{18}\) + \(\dfrac{15}{18}\) - \(\dfrac{63}{18}\))
= \(\dfrac{3}{5}\) \(\times\) (- \(\dfrac{23}{9}\))
= - \(\dfrac{23}{15}\)
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Tính nhanh ( Giúp mình nha )
a) -3/7 + 15/26 - ( 2/13 - 3/7 )
b) 2.3/7 + (2/9 -1 3?7 ) -5/3 :1/9
c) -11/23* 6/7 +8/7 * -11/23 - 1/23
d)(377/-231-123/89+ 34/791) * (1/6 -1/8-1/24)
- Kasamita :)))
CMR : 3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 + ... + 19/9^2.10^2 < 1
Ta có :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=1-\frac{1}{10^2}< 1\)
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2 tháng 7 2021 lúc 8:43
Chứng minh rằng :
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
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3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 +...+ 19/9^2.10^2. chung minh nho hon 1
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\left(\frac{1}{3^2}-\frac{1}{4^2}\right)+...+\left(\frac{1}{9^2}-\frac{1}{10^2}\right)\)
\(=\frac{1}{1}-\frac{1}{10^2}\)
\(=1-\frac{1}{100}
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=3/1.4+5/4.9+7/9.16+......+19/81.100
=(1/1-1/4)+(1/4-1/9)+........+(1/81-1/100)
=1-1/100
=99/100<1(đpcm)
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Xem thêm câu trả lời
Chứng minh rằng :
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\)
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