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a) 11/12 x 9/19 - 22/24 x 6/19 + 11/12 x 16/19.
= 11/12 x 9/19 - 11/12 x 6/19 + 11/12 x 16/19.
=11/12 x ( 9/19 -6/19 + 16/19)
=11/12x 1
=11/12
mình biết câu a/ thôi
Bài 1:
\(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - \(\dfrac{2}{5}\)
=\(\dfrac{40}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{61}{60}\)
b; \(\dfrac{12}{13}\) x \(\dfrac{3}{4}\) - \(\dfrac{7}{13}\)
= \(\dfrac{9}{13}\) - \(\dfrac{7}{13}\)
= \(\dfrac{2}{13}\)
c; \(\dfrac{15}{17}\) : \(\dfrac{19}{34}\) - \(\dfrac{17}{19}\)
= \(\dfrac{15}{17}\) x \(\dfrac{34}{19}\) - \(\dfrac{17}{19}\)
= \(\dfrac{30}{19}\) - \(\dfrac{17}{19}\)
= \(\dfrac{13}{19}\)
Bài 2:
a; \(\dfrac{x}{5}\) x \(\dfrac{3}{7}\) = \(\dfrac{9}{35}\)
\(\dfrac{x}{5}\) = \(\dfrac{9}{35}\) : \(\dfrac{3}{7}\)
\(\dfrac{x}{5}\) = \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{3}{5}\) x 5
\(x\) = 3
a, \(\frac{5}{11}\times\frac{7}{25}+\frac{15}{11}\times\frac{1}{5}\)
\(=\frac{5}{11}\times\frac{7}{25}+\frac{5}{11}\times\frac{3}{5}\)
\(=\frac{5}{11}\times\left(\frac{7}{25}+\frac{3}{5}\right)\)
\(=\frac{5}{11}\times\frac{22}{25}\)
\(=\frac{2}{5}\)
b, \(\frac{3}{7}\times\frac{25}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\frac{75}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\left(\frac{75}{19}-\frac{18}{19}\right)\)
\(=\frac{1}{7}\times3\)
\(=\frac{3}{7}\)
1) \(=\left(\frac{3}{5}+\frac{2}{5}\right).\frac{6}{11}\)
\(=1.\frac{6}{11}\)
\(=\frac{6}{11}\)
2)\(\frac{17}{25}.\left(\frac{11}{19}+\frac{6}{19}+\frac{2}{19}\right)\)
\(=\frac{17}{25}.1\)
\(=\frac{17}{25}\)
Tích có các thừa số là: (199-19):10+1=19(số)
Vì 19x29= TC1 (TC1 là tận cùng =1)
có số cặp là: 19:2=9 Dư 1 thừa số
Ta có: TC1xTC1xTC1x...TC1x TC9=TC9
Bài 1:
Rút gọn các phân số sau:
\(\dfrac{12}{24}\) = \(\dfrac{12:12}{24:12}\) = \(\dfrac{1}{2}\)
\(\dfrac{120}{90}\) = \(\dfrac{120:30}{90:30}\) = \(\dfrac{4}{3}\)
\(\dfrac{42}{24}\) = \(\dfrac{42:6}{24:6}\) = \(\dfrac{7}{4}\)
Bài 2:
Tính bằng cách thuận tiện:
\(\dfrac{13}{19}\) + \(\dfrac{18}{19}\) + \(\dfrac{17}{19}\)
= \(\dfrac{13+18+17}{19}\)
= \(\dfrac{\left(13+17\right)+18}{19}\)
= \(\dfrac{48}{19}\)
\(\dfrac{12}{17}\) + \(\dfrac{19}{17}\) + \(\dfrac{8}{17}\)
= \(\dfrac{12+19+8}{17}\)
= \(\dfrac{\left(12+8\right)+19}{17}\)
= \(\dfrac{39}{17}\)
a) \(D=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...++\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
=> \(D=1-\frac{1}{1024}\)
b) \(Đ=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
a) D=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{512}+\frac{1}{1024}.\)
\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\dots+\frac{1}{512}-\frac{1}{1024}\)
\(D=1-\frac{1}{1024}\)
\(D=\frac{1023}{1024}\)
\(Đ=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(Đ=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(Đ=1-\frac{1}{20}\)
\(Đ=\frac{19}{20}\)
Phần c như kiểu sai đề chỗ cuối hay sao ấy.
\(\frac{24}{38}+x=\frac{21}{19}-\frac{5}{19}\)
\(\Rightarrow\frac{24}{38}+x=\frac{16}{19}\)
\(\Rightarrow x=\frac{16}{19}-\frac{24}{38}\)
\(\Rightarrow x=\frac{16}{19}-\frac{12}{19}\)
\(\Rightarrow x=\frac{4}{19}\)
Bài 2 tự làm
đáp số: 62868
62868