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a, \(\frac{2020.125+1000}{126.2020-1020}=\frac{2020.125+1000}{2020.125+2020-1020}=1\)
b,\(\left(\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{19.21}\right).426+x=19\)
\(< =>\left(\frac{1}{11}-\frac{1}{21}\right).213+x=19\)
\(< =>\frac{2130}{231}+x=19\)
\(< =>x=19-\frac{2130}{231}=...\)
a)\(\frac{2020\cdot125+1000}{126\cdot2020-1020}=\frac{2020\cdot125+1000}{2020\cdot125+2020-1020}=\frac{2020\cdot125+1000}{2020\cdot125+1000}=1\)
b) \(\left(\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}+\frac{1}{19\cdot21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{11}-\frac{1}{21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{10}{231}\cdot426+x=19\)
\(\Leftrightarrow\frac{710}{77}+x=19\)
\(\Leftrightarrow x=19-\frac{710}{77}=\frac{753}{77}\)
a. nhân cả hai vế của đẳng thức với 1/ 10 ta có
x/10 - (2/11.13 +2/13.15+...+2/53.55)=3/11 . 1/10
x/10 - (1/11-1/13+1/13-1/15 +...+1/53-1/55) =3/110
x/10 - (1/11 - 1/55) =3/110
x/10 -4/55 = 3/110
x/10=3/110 + 4/55
x. 1/10 =1/10
x= 1/10 : 1/10 =1
b) bạn nhân cả hai vế của đẳng thức với 1/2 rồi làm tương tự
a. nhân cả hai vế của đẳng thức với \(\frac{1}{10}\). Ta có:
\(\frac{x}{10}-\left(\frac{2}{11.13}+\frac{2}{13.15}+...\frac{2}{53.55}\right)=\frac{3}{11}.\frac{1}{10}\)
\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{110}\)
\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{110}\)
\(\frac{x}{10}-\frac{-4}{55}=\frac{3}{110}\)
\(\frac{x}{10}=\frac{3}{110}+\frac{4}{55}\)
\(x.\frac{1}{10}=\frac{1}{10}\)
\(x=\frac{1}{10}:\frac{1}{10}=1\)
b. cũng thế bạn nhân hai vế của đẳng thức với \(\frac{1}{2}\) rồi làm tương tự.
x-10( 2/11.13+2/13.15+...+2/53.55)=3/11
x-10(1/11-1/55)=3/11
x-10.4/55=3/11
x-40/55=3/11
x=3/11+40/55
x= 1
x-10.(2/11.13+2/13.15+....+2/53.55)=3/11
x-10.(1/11-1/13+...+1/53-1/55)=3/11
x-10.(1/11-1/55)=3/11
x-10.4/55=3/11
x-8/11=3/11
x=1
Nhớ k cho mình nhé
\(M=\frac{3}{2}.\left(\frac{2}{11.13}+\frac{2}{13.15}+......+\frac{2}{97.99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+.....+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{99}\right)=\frac{3}{2}.\frac{8}{99}=\frac{4}{33}\)
M= \(\frac{3}{11\cdot13}+\frac{3}{13\cdot15}+\frac{3}{15\cdot17}+...+\frac{3}{97\cdot99}\)
=\(\frac{3}{2}\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+...+\frac{2}{97\cdot99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\frac{8}{99}\)
= \(\frac{4}{33}\)
M=\(3\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+....+\frac{1}{97}-\frac{1}{99}\right)\)
M=3(\(\frac{1}{11}-\frac{1}{99}\))
M=3.8/99
M=8/33
M=3/11.13 + 3/13.15 + 3/15.17 + .... + 3/97.99
M=3/2 .(2/11.13 + 2/13.15 + 2/15.17 +... + 2/97.99)
M=3/2.(1/11-1/13+1/13-1/15+1/15-1/17+...+1/97-1/99)
M=3/2.(1/11-1/99)
M=3/2.(9/99-1/99)
M=3/2.8/99
M=4/33
Vậy M =4/33
\(M=\frac{3}{11.13}+\frac{3}{13.15}+\frac{3}{15.17}+...+\frac{3}{97.99}\)
\(M=\frac{3}{2}\times\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{97.99}\right)\)
\(M=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(M=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(M=\frac{3}{2}\times\frac{8}{99}\)
\(M=\frac{4}{33}\)
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