(0,5 điểm) Tính $\dfrac{-3}{11}-0,251-\dfrac{8}{11}+2,251$.
(0,5 điểm) Tính.
$\dfrac{11}{3} \cdot \dfrac{2}{5}+\dfrac{11}{3} \cdot \dfrac{8}{5}-\dfrac{11}{3}$
11/3 . 2/5 + 11/3 . 8/5 - 11/3
= 11/3 . (2/5 + 8/5 - 1)
= 11/3 . (2 - 1)
= 11/3 . 1
= 11/3
\(\dfrac{11}{3}\)⋅\(\dfrac{2}{5}\)+\(\dfrac{11}{3}\).\(\dfrac{8}{5}\)-\(\dfrac{11}{3}\)
= \(\dfrac{11}{3}\).(\(\dfrac{2}{5}\) + \(\dfrac{8}{5}\) - 1)
= \(\dfrac{11}{3}\).(2 - 1)
= \(\dfrac{11}{3}\) . 1
=\(\dfrac{11}{3}\)
Bài 1 . Thực hiện phép tính sau :
a. \(\dfrac{-4}{11}\). \(\dfrac{7}{9}\)+\(\dfrac{-4}{11}\) . \(\dfrac{2}{9}\) - \(\dfrac{7}{11}\)
b. \(\dfrac{3}{5}\):\(\dfrac{-7}{10}\)+ 0,5 - \(\left(-\dfrac{9}{14}\right)\)
c. \(\dfrac{3}{5}\)-\(\dfrac{8}{5}\): (5,25+75%)
Bài 2 . Tìm x :
a) x - \(\dfrac{1}{2}\)= \(\dfrac{-1}{10}\) b.\(\dfrac{2}{3}x\) - \(\dfrac{7}{6}\)=\(\dfrac{5}{2}\) c)2,5 - \(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)\) =\(\dfrac{3}{4}\)
Mik làm Bài 2 nhé ~
Bài 2 :
a) \(x-\dfrac{1}{2}=-\dfrac{1}{10}\)
\(x=-\dfrac{1}{10}+\dfrac{1}{2}\)
\(x=\dfrac{2}{5}\)
b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\)
\(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\)
\(\dfrac{2}{3}x=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}:\dfrac{2}{3}\)
\(x=\dfrac{11}{3}.\dfrac{3}{2}\)
\(x=\dfrac{11}{2}\)
c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=2,5-\dfrac{3}{4}\)
\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{5}{2}-\dfrac{3}{4}\)
\(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\dfrac{1}{8}x=\dfrac{5}{4}\)
\(x=10\)
Bài 1:
a) \(\dfrac{-4}{11}.\dfrac{7}{9}+\dfrac{-4}{11}.\dfrac{2}{9}-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}.\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}.1-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}-\dfrac{7}{11}\)
\(=-1\)
b) \(\dfrac{3}{5}:\dfrac{-7}{10}+0,5-\left(\dfrac{-9}{14}\right)\)
\(=\dfrac{-6}{7}+\dfrac{1}{2}+\dfrac{9}{14}\)
\(=\dfrac{2}{7}\)
c) \(\dfrac{3}{5}-\dfrac{8}{5}:\left(5,25+75\%\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:\left(\dfrac{21}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:6\)
\(=\dfrac{3}{5}-\dfrac{4}{15}\)
\(=\dfrac{1}{3}\)
Bài 2:
a) \(x-\dfrac{1}{2}=\dfrac{-1}{10}\)
\(x=\dfrac{-1}{10}+\dfrac{1}{2}\)
\(x=\dfrac{2}{5}\)
b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\)
\(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\)
\(\dfrac{2}{3}x=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}:\dfrac{2}{3}\)
\(x=\dfrac{11}{2}\)
c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{5}{2}-\dfrac{3}{4}\)
\(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\dfrac{1}{8}x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}:\dfrac{1}{8}\)
\(x=10\)
(0,5 điểm) Tính:
$\sqrt{25}.\left( 0,4-1\dfrac{1}{2} \right) : \left[ (-2)^3 : \dfrac{8}{11} \right]$.
\(\sqrt{25}.\left(0,4-1\dfrac{1}{2}\right):\left[\left(-2\right)^3:\dfrac{8}{11}\right]\)
\(=5.\left(\dfrac{2}{5}-\dfrac{3}{2}\right):\left(-8:\dfrac{8}{11}\right)\)
\(=5.\left(-\dfrac{11}{10}\right):\left(-11\right)\)
\(=\dfrac{-11}{2}:\left(-11\right)\)
\(=\dfrac{1}{2}\)
\(5\).\(\left(0,4-1,5\right):[-8:\dfrac{8}{11}]\)
=5.-11:-11
5
Tính một cách hợp lí:
\(A = \left( { - \dfrac{3}{{11}}} \right) + \dfrac{{11}}{8} - \dfrac{3}{8} + \left( { - \dfrac{8}{{11}}} \right)\)
\(\begin{array}{l}A = \left( { - \dfrac{3}{{11}}} \right) + \dfrac{{11}}{8} - \dfrac{3}{8} + \left( { - \dfrac{8}{{11}}} \right)\\ = \left[ {\left( { - \dfrac{3}{{11}}} \right) + \left( { - \dfrac{8}{{11}}} \right)} \right] + \left( {\dfrac{{11}}{8} - \dfrac{3}{8}} \right)\\ = \dfrac{{ - 11}}{{11}} + \dfrac{8}{8} = - 1 + 1 = 0\end{array}\)
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,265+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\)
Thực hiện phép tính: (2 điểm)
a) $1-\dfrac{1}{2}+\dfrac{1}{3}$ ;
b) $\dfrac{2}{5}+\dfrac{3}{5}: \dfrac{9}{10}$ ;
c) $\dfrac{7}{11} \cdot \dfrac{3}{4}+\dfrac{7}{11} \cdot \dfrac{1}{4}+\dfrac{4}{11}$ ;
d) $\left(\dfrac{3}{4}+0,5+25 \%\right) \cdot 2 \dfrac{2}{3}$ .
a) = 1/2 + 1/3 = 5/6
b) = 2/5 + 2/3 = 16/15
c) = 7/11 . ( 3/4 + 1/4 ) + 4/11
= 7/11 . 1 + 4/11
= 7/11 + 4/11
= 1
d) = ( 3/4 + 1/2 + 1/4 ) . 8/3
= 3/2 . 8/3
= 4
Câu 1.
a) 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)= \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)=\(\dfrac{3}{6}\)+\(\dfrac{2}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{2}{5}\)+\(\dfrac{3}{5}\):\(\dfrac{9}{10}\)= 1 :\(\dfrac{9}{10}\)= \(\dfrac{10}{9}\)
c) \(\dfrac{7}{11}\).\(\dfrac{3}{4}\)+\(\dfrac{7}{11}\).\(\dfrac{1}{4}\)+\(\dfrac{4}{11}\)=\(\dfrac{7}{11}\).1+\(\dfrac{4}{11}\)=1
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-265+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\)+\(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(\dfrac{0.375-0.3+\dfrac{3}{11}+\dfrac{3}{12}}{-0.625+0.5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1.5+1-0.75}{2.5+\dfrac{5}{3}-1.25}\)
=\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
=\(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
=\(\dfrac{3}{-5}+\dfrac{3}{5}\)
=\(0\)
B = \(\dfrac{5}{2}\)+\(\dfrac{6}{11}\)+\(\dfrac{3}{8}\)+\(\dfrac{7}{2}\)+\(\dfrac{6}{8}\)+\(\dfrac{5}{11}\)
tính nhanh
Mình nghĩ đề là : 2/8 sẽ hay hơn.
\(B=\dfrac{5}{2}+\dfrac{6}{11}+\dfrac{2}{8}+\dfrac{7}{2}+\dfrac{6}{8}+\dfrac{5}{11}\)
\(=\left(\dfrac{5}{2}+\dfrac{7}{2}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\left(\dfrac{2}{8}+\dfrac{6}{8}\right)\)
\(=6+1+1=8\)
\(B=\dfrac{5}{2}+\dfrac{6}{11}+\dfrac{3}{8}+\dfrac{7}{2}+\dfrac{6}{8}+\dfrac{5}{11}\)
\(B=\left(\dfrac{5}{2}+\dfrac{7}{2}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\left(\dfrac{3}{8}+\dfrac{6}{8}\right)\)
\(B=6+1+1,125\)
\(B=8,125\)
Helpppppppppppp
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)