Bài 1 Rút gọn a
\(\frac{1919}{2121}\)
b \(\frac{191191}{212212}\)
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 6
a; \(\dfrac{1919}{2121}\) = \(\dfrac{1919:101}{2121:101}\) = \(\dfrac{19}{21}\)
b; 17 x 5 - \(\dfrac{17}{120}\) - 103
= 85 - \(\dfrac{17}{120}\) - 103
= \(\dfrac{10200}{120}\) - \(\dfrac{17}{120}\) - \(\dfrac{12360}{120}\)
= \(\dfrac{10183}{120}\) - \(\dfrac{12360}{120}\)
= \(\dfrac{-2177}{120}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 6
c; 2929 - \(\dfrac{101}{2\times1919}\) + 404
= 2929 - \(\dfrac{1}{38}\) + 404
= \(\dfrac{111302}{38}\) - \(\dfrac{1}{38}\) + \(\dfrac{15352}{38}\)
= \(\dfrac{111301}{38}\) + \(\dfrac{15352}{38}\)
= \(\dfrac{126653}{38}\)
TM
6
14 tháng 8 2018
18 x \(\left(\frac{1919}{2121}+\frac{888}{999}\right)\)
= 18 x \(\frac{113}{63}\)
= \(\frac{226}{7}\)
TM
4
14 tháng 8 2018
\(18.\frac{113}{63}\)
\(=\frac{226}{7}\)
k mk nha mk nhanh nhất
N
18 tháng 7 2017
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{7}{15}\)
TN
18 tháng 7 2017
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)=\frac{1}{2}.\frac{14}{15}\)\(=\frac{7}{15}\)
b)\(\frac{1414+1515+...+1919}{2020+2121+...+2525}\)
\(\Rightarrow\frac{101\left(14+15+16+17+18+19\right)}{101\left(20+21+22+23+24+25\right)}\)
\(=\frac{14+15+16+17+18+19}{20+21+22+23+24+25}\)
\(=\frac{\left(19+14\right).6:2}{\left(25+20\right).6:2}=\frac{19+14}{25+20}=\frac{33}{45}=\frac{11}{15}\)
DG
1
DN
3
25 tháng 1 2019
1.
a + b + c = 0 \(\Rightarrow\)a = - ( b + c ) \(\Rightarrow\)a2 = [ -( b + c ) ]2 \(\Rightarrow\)a2 = b2 + c2 + 2bc
Tương tự : b2 = a2 + c2 + 2ac ; c2 = a2 + b2 + 2ab
a + b + c = 0 \(\Rightarrow\)a3 + b3 + c3 = 3abc ( chứng minh )
Ta có : \(A=\frac{a^2}{b^2+c^2+2bc-b^2-c^2}+\frac{b^2}{a^2+c^2+2ac-a^2-c^2}+\frac{c^2}{a^2+b^2+2ab-a^2-b^2}\)
\(A=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\)
\(A=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
2. quy đồng mà giải
1919/2121 = 19/21
191191/212212=191/212
a ) \(\frac{1919}{2121}=\frac{19.101}{21.101}=\frac{19}{21}\)
b) \(\frac{191191}{212212}=\frac{191.1001}{212.1001}=\frac{191}{212}\)