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=\(\left(13\frac{9}{11}.\frac{49}{38}-5\frac{2}{11}.\frac{49}{38}\right).\left(\frac{38}{49}.\frac{11}{5}\right)\)
= \(\left(13\frac{9}{11}-5\frac{2}{11}\right).\frac{49}{38}.\frac{38}{49}.\frac{11}{5}\) = \(\left(13+\frac{9}{11}-5-\frac{2}{11}\right).\frac{11}{5}\)
= \(\left(8+\frac{7}{11}\right).\frac{11}{5}\)= \(\frac{95}{11}.\frac{11}{5}=19\)
\(\left(13\frac{9}{11}:\frac{38}{49}-5\frac{2}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)
\(=\left(\frac{152}{11}:\frac{38}{49}-\frac{57}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)
\(=\left(\frac{196}{11}-\frac{147}{22}\right):\frac{245}{418}=\frac{245}{22}:\frac{245}{418}=19\)
Mình làm thử, không biết có sai không, nếu sai thì cho mình đáp án rút kinh nghiệm nhé:
\(\left(13\frac{9}{11}:\frac{38}{49}-5\frac{2}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)
\(=\left(\frac{152}{11}:\frac{38}{49}-\frac{57}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)
\(=\left(\frac{196}{11}-\frac{147}{22}\right):\frac{245}{418}\)
\(=\frac{245}{22}:\frac{245}{418}=19\)
\(3\frac{1}{2}.\frac{4}{49}-\left[2,\left(4\right).2\frac{5}{11}\right]:\left(\frac{-42}{5}\right)\)
\(3\frac{1}{2}.\frac{4}{49}-\left[2,\left(4\right).2\frac{5}{11}\right]:\left(-\frac{42}{5}\right)\)
\(=\frac{7}{2}.\frac{4}{49}-\left[\frac{22}{9}.\frac{27}{11}\right]:\left(-\frac{42}{5}\right)\)
\(=\frac{7}{2}.\frac{4}{49}-6:\left(-\frac{42}{5}\right)\)
\(=\frac{2}{7}-\left(-\frac{5}{7}\right)\)
\(=\frac{2}{7}+\frac{5}{7}\)
\(=1.\)
Chúc bạn học tốt!
\(0,4\left(3\right)+0,\left(6\right)2.2\frac{1}{2}-\frac{\frac{1}{2}+\frac{1}{3}}{0,\left(5\right)8}:\frac{50}{53}\)
Giải hộ mik câu này nx nha
a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}=\left(\frac{11}{24}+\frac{13}{24}\right)+\left(-\frac{5}{41}-\frac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
_Học tốt nha_
a, \(\frac{15}{12}\)+ \(\frac{5}{13}\)- \(\frac{3}{12}\)-\(\frac{18}{13}\)
= \(\frac{5}{4}\)+ \(\frac{5}{13}\) - \(\frac{1}{4}\) - \(\frac{18}{13}\)
= \(\left(\frac{5}{4}-\frac{1}{4}\right)\)+ \(\left(\frac{5}{13}-\frac{18}{13}\right)\)
= 1 - 1 = 0
b, \(\frac{11}{24}\)- \(\frac{5}{41}\)+ \(\frac{13}{24}\)+ 0,5 - \(\frac{36}{41}\)
= \(\left(\frac{11}{24}+\frac{13}{24}\right)\)- \(\left(\frac{5}{41}+\frac{36}{41}\right)\)+ 0,5
= 1 - 1 + 0,5 = 0,5
c, \(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
=\(\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{5}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{5}{11}\)
= \(\frac{11}{5}.\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)
= \(\frac{11}{5}.\left[\left(-\frac{3}{4}-\frac{1}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)\right]\)
= \(\frac{11}{5}.\left[\left(-1\right)+1\right]\)
= 0
d, \(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
= \(9.\left(0,75-0,25\right)-2\)
= 9. 0,5 - 2 = 2,5
e, \(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= -1 + 1 - \(\frac{1}{2}\)
= \(-\frac{1}{2}\)
Xét dãy tích P1 ta thấy 2 thừa số đều âm
=> P1 dương <=> P1 > 0
Xét dãy tích P2 ta thấy có 3 thừa số âm
=> P2 âm <=> P2 < 0
XXets dãy P3 thấy trong đó có một thừa số là \(\frac{0}{11}=0\)
=> P3 = 0
Vậy P2 < P3 < P1
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)