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\(=\dfrac{7}{8}\left(\dfrac{2}{9}-\dfrac{1}{18}+\dfrac{1}{36}-\dfrac{5}{12}\right)=\dfrac{7}{8}\cdot\dfrac{8-2+1-15}{36}\)
\(=\dfrac{7}{8}\cdot\dfrac{-8}{36}=\dfrac{-7}{36}\)
a) \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
⇔ \(\left(\dfrac{-8}{5}+x\right).\dfrac{13}{12}=\dfrac{13}{6}\)
⇔ \(-\dfrac{8}{5}+x=\dfrac{13}{6}:\dfrac{13}{12}\)
⇔ \(-\dfrac{8}{5}+x=2\)
⇔ \(x=2+\dfrac{8}{5}\)
⇔ \(x=\dfrac{18}{5}\)
b) \(\dfrac{-4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
⇔ \(-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
⇔ \(-\dfrac{4}{7}x=-\dfrac{3}{40}-\dfrac{7}{5}\)
⇔ \(-\dfrac{4}{7}x=-\dfrac{59}{40}\)
⇔ \(x=\left(-\dfrac{59}{40}\right):\left(-\dfrac{4}{7}\right)\)
⇔ \(x=\dfrac{413}{160}\)
a, \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
=> \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=\dfrac{13}{6}\)
=> \(\left(-1\dfrac{3}{5}+x\right)=\dfrac{13}{6}.\dfrac{12}{13}\)
=> \(\left(-1\dfrac{3}{5}+x\right)=2\)
=> \(\dfrac{-8}{5}+x=2\)
=> x= \(2+\dfrac{8}{5}=\dfrac{10}{5}+\dfrac{8}{5}\)
=> x= \(\dfrac{18}{5}\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
\(\left(-\dfrac{4}{3}+\dfrac{5}{13}\right):\dfrac{2}{7}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right):\dfrac{2}{7}\\ =\left(-\dfrac{4}{3}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right):\dfrac{2}{7}\\ =-\dfrac{595}{156}:\dfrac{2}{7}\\ =-\dfrac{595}{156}.\dfrac{7}{2}=-\dfrac{4165}{312}\)
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
Lời giải:
a)
\(\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}}{\frac{8}{3}-\frac{8}{5}+\frac{8}{7}-\frac{8}{9}+\frac{8}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}\) \(=\frac{2}{8}=\frac{1}{4}\)
b)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)
\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(A=\left(-\dfrac{5}{7}+\dfrac{8}{5}\right):\dfrac{91}{8}+\left(-\dfrac{2}{7}-\dfrac{3}{5}\right):\dfrac{91}{8}\)
\(=\dfrac{31}{35}:\dfrac{91}{8}+\dfrac{-31}{35}:\dfrac{91}{8}\)
\(=\dfrac{248}{3185}+\dfrac{-248}{3185}\)
= 0
\(B=\dfrac{13}{15}:\left(\dfrac{4}{5}-\dfrac{3}{7}\right)+\dfrac{13}{15}:\left(\dfrac{2}{5}-\dfrac{1}{9}\right)\)
\(=\dfrac{13}{15}:\dfrac{13}{35}+\dfrac{13}{15}:\dfrac{13}{45}\)
\(=\dfrac{7}{3}+3\)
\(=\dfrac{16}{3}\)
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
\(\dfrac{7}{8}\): \(\left(\dfrac{2}{9}-\dfrac{1}{18}\right)\)+\(\dfrac{7}{8}\): \(\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
= \(\dfrac{7}{8}\): \(\dfrac{1}{6}\)+\(\dfrac{7}{8}\): \(\left(\dfrac{-14}{36}\right)\)
= \(\dfrac{7}{8}\): \(\left\{\dfrac{1}{6}+\left(\dfrac{-14}{36}\right)\right\}\)
= \(\dfrac{7}{8}\): \(\left(\dfrac{-2}{9}\right)\)
= \(\dfrac{7}{8}\). \(\left(\dfrac{-9}{2}\right)\)
= \(\dfrac{-63}{16}\)
Giải:
\(\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{8}:\left(\dfrac{4}{18}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{15}{36}\right)\)
\(=\dfrac{7}{8}:\dfrac{3}{18}+\dfrac{7}{8}:\left(-\dfrac{7}{18}\right)\)
\(=\dfrac{7}{8}.6+\dfrac{7}{8}.\dfrac{18}{-7}\)
\(=\dfrac{7}{8}\left(6+\dfrac{18}{-7}\right)\)
\(=\dfrac{7}{8}.\dfrac{24}{7}\)
\(=3\)
Vậy giá trị của biểu thức trên là 3.
Chúc bạn học tốt!!!