Tính hợp lý.
\(a0.5+\frac{1}{3}+0.4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
\(b\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
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\(\frac{9}{8}-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{72}=\frac{9}{8}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)
\(=\frac{9}{8}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
\(=\frac{9}{8}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{9}{8}-\left(1-\frac{1}{9}\right)=\frac{9}{8}-\frac{8}{9}=\frac{17}{72}\)
Đầu tiên , ta cộng các phần nguyên lại với nhau trước :
( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{8}{72}+\frac{1}{90}+\frac{1}{10}\)
= 45 + \(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{42}+\frac{1}{72}\right)+\left(\frac{1}{10}+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{90}\right)+\frac{1}{56}\)
= 45 +
tới đây tớ chịu , các cậu giúp với
Đầu tiên , cộng các phần nguyên lại với nhau , ta có :
( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + ( \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\))
= 45 + \(\left(\frac{1}{6}+\frac{1}{30}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc , ta được 6 / 30 , rút gọn tối giản còn 1 / 5
= 45 + \(\left(\frac{1}{5}+\frac{1}{20}\right)+\frac{1}{2}+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc và rút gọn tối giản , ta được 1 / 4
= 45 + \(\left(\frac{1}{4}+\frac{1}{2}\right)+\frac{1}{12}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi cộng trong ngoặc rồi rút gọn , ta được 3 / 4
= 45 + \(\left(\frac{3}{4}+\frac{1}{12}\right)+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
rút gọn lại ta được 5 / 6
= 45 + \(\left(\frac{5}{6}+\frac{1}{42}\right)+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
rút gọn tối giản ra 6 / 7
= 45 + \(\left(\frac{6}{7}+\frac{1}{56}\right)+\frac{1}{72}+\frac{1}{90}+\frac{1}{10}\)
sau khi tính trong ngoặc rút gọn được 7 / 8
= 45 + \(\left(\frac{7}{8}+\frac{1}{72}\right)+\frac{1}{90}+\frac{1}{10}\)
tính trong ngoặc rồi rút gọn ra 8 / 9
= 45 + \(\left(\frac{8}{9}+\frac{1}{90}\right)+\frac{1}{10}\)
cũng rút gọn tiếp ta được 9 / 10
= 45 + \(\left(\frac{9}{10}+\frac{1}{10}\right)\)
= 45 + 1
= 46
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+.....+\frac{1}{2}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+........+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}...-\frac{1}{6}-\frac{1}{2}\)
= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)
= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
= \(\frac{8}{9}-\frac{8}{9}\)
= \(0\)
"girl cute" là sai rồi bạn ơi, trong tiếng anh, tính từ (cute) phải đứng trước danh tư (girl).
a, \(A=\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{8}{9}-\left(\frac{1}{8}-\frac{1}{9}+\frac{1}{7}-\frac{1}{8}+\frac{1}{6}-\frac{1}{7}+...+1-\frac{1}{2}\right)\)
\(A=\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
\(A=\frac{8}{9}-\frac{8}{9}\)
\(A=0\)
\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)
\(S=\frac{3}{4}-\frac{1}{4}-\left[\frac{14}{6}+\left(\frac{-27}{6}\right)\right]-\frac{5}{6}\)
\(S=\frac{1}{2}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)
\(S=\frac{3}{6}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)
\(S=\frac{11}{6}\)
\(D=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
Làm tắt nha :
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(D=\frac{1}{2}.\frac{98}{99}-\frac{1}{2}.\frac{98}{100}\)
\(D=\frac{1}{2}\left(\frac{98}{99}-\frac{98}{100}\right)\)
Tự tính nốt nha