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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
a, \(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right)-\dfrac{5}{21}:\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\\ = -\dfrac{5}{21}:\dfrac{4}{5}+ \left(-\dfrac{5}{21}\right):\dfrac{4}{5}\\ =\left[-\dfrac{5}{21}+\left(-\dfrac{5}{21}\right)\right]:\dfrac{4}{5}\\ -\dfrac{10}{21}:\dfrac{4}{5}\\ =-\dfrac{25}{42}\)
b,
\(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\\ =\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:-\dfrac{3}{5}\\ =\dfrac{5}{9}:\left(\dfrac{-3}{22}+-\dfrac{3}{5}\right)\\ =\dfrac{5}{9}:-\dfrac{81}{110}\\ =-\dfrac{550}{729}\)
b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)
\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)
c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)
\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)
hay \(x=-\dfrac{26}{21}\)
a: \(\Leftrightarrow\left(3x-2\right):\dfrac{7}{5}=\dfrac{17}{7}:\dfrac{13}{5}=\dfrac{85}{91}\)
\(\Leftrightarrow3x-2=\dfrac{85}{91}\cdot\dfrac{7}{5}=\dfrac{17}{13}\)
=>3x=43/13
hay x=43/39
b: \(\Leftrightarrow9x+207=121-8x\)
=>19x=-86
hay x=-86/19
c: \(\Leftrightarrow x^2-9=16\)
=>x2=25
=>x=5 hoặc x=-5
d: \(\Leftrightarrow\left|x\right|=\dfrac{1.64\cdot3.11}{8.51}\simeq0,6\)
=>x=0,6 hoặc x=-0,6
\(A=\dfrac{5}{4}\cdot\dfrac{15-4}{3}\cdot\dfrac{-1}{11}=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}\)=-50/120
\(B=\dfrac{3}{4}\cdot\dfrac{-1}{12}\cdot\dfrac{-2}{3}=\dfrac{3\cdot2}{4\cdot12\cdot3}=\dfrac{2}{4\cdot12}=\dfrac{1}{24}\)=5/120
\(C=\dfrac{5}{4}\cdot\dfrac{-1}{15}\cdot\dfrac{-2}{5}=\dfrac{2}{4\cdot15}=\dfrac{1}{30}\)=4/120
\(D=3\cdot\dfrac{8-15}{12}\cdot\dfrac{-1}{7}=\dfrac{1}{4}\)=30/120
Vì -50<4<5<30
nên A<C<B<D
a, \(\left(0,8.7+0,64\right).\left(1,25.7-\dfrac{4}{5}.1,25\right)\)
\(=6,24.1,25\left(7-\dfrac{4}{5}\right)=7,8.6,2=48,36\)
b, \(\dfrac{1}{4}-\dfrac{\dfrac{1}{4}+\dfrac{1}{9}}{\dfrac{1}{9}}=\dfrac{1}{4}-\dfrac{\dfrac{13}{36}}{\dfrac{1}{9}}=\dfrac{1}{4}-\dfrac{13}{4}=-3\)
c, \(2\dfrac{1}{2}-3,4\left(12\right)-\dfrac{4}{3}+\dfrac{1}{3}.\left(\dfrac{1}{2}+0,5-3\dfrac{1}{2}\right)\)
\(=\dfrac{5}{2}-\dfrac{563}{165}-\dfrac{4}{3}+\dfrac{1}{3}.\left(-\dfrac{5}{2}\right)\)
\(=-\dfrac{301}{330}-\dfrac{4}{3}-\dfrac{5}{6}=-\dfrac{508}{165}\)
Chúc bạn học tốt!!!
a. \(\dfrac{5}{3}+\left(\dfrac{-2}{7}\right)-\left(-1.2\right)\)
= \(\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\)
= \(\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}\)
= \(\dfrac{271}{105}\)
b.\(\dfrac{-4}{7}+\left(\dfrac{-5}{6}\right)-\dfrac{17}{4}\)
= \(\dfrac{-4}{9}-\dfrac{5}{6}-\dfrac{17}{4}\)
= \(\dfrac{-16}{36}-\dfrac{30}{36}-\dfrac{153}{36}\)
= \(\dfrac{-199}{36}\)
a. \(\dfrac{3}{4}-\left(2x-\dfrac{2}{3}\right)=\dfrac{-5}{6}\)
\(\Rightarrow2x-\dfrac{2}{3}=\dfrac{3}{4}-\dfrac{-5}{6}\)
\(\Rightarrow2x-\dfrac{2}{3}=\dfrac{19}{12}\)
\(\Rightarrow2x=\dfrac{19}{12}+\dfrac{2}{3}=\dfrac{9}{4}\)
\(\Rightarrow x=\dfrac{9}{4}:2=\dfrac{9}{8}\)
Vậy............
b. \(1,5-\left(x+\dfrac{7}{2}\right)=2^7:2^5\)
\(\Rightarrow1,5-\left(x+\dfrac{7}{2}\right)=2^2=4\)
\(\Rightarrow x+\dfrac{7}{2}=1,5-4=\dfrac{-5}{2}\)
\(\Rightarrow x=\dfrac{-5}{2}-\dfrac{7}{2}=-6\)
Vậy.............
a,\(\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}\)
b,\(\dfrac{-4}{3}:\dfrac{2}{15}=\dfrac{-60}{6}=-10\)
c,\(\dfrac{3}{7}.\dfrac{2}{9}+\dfrac{7}{9}.\dfrac{3}{7}=\dfrac{3}{7}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{3}{7}\)
d,\(\left(\dfrac{-2}{3}\right)^3.9^2+\left(\dfrac{-3}{4}\right)^2.32\)
\(=\dfrac{\left(-2\right)^3}{3^3}.3^4+\dfrac{\left(-3\right)^2}{4^2}.2^5\)
\(=\left(-8\right).3+\dfrac{3^2}{4^2}.2^5\)
\(=\left(-24\right)+2.9\)
\(=\left(-24\right)+18\)
\(=-6\)
a,