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a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)=\dfrac{7}{8}.\left(\dfrac{10}{60}+\dfrac{24}{60}\right)=\dfrac{7}{8}.\dfrac{17}{30}=\dfrac{114}{240}\)
b)\(\dfrac{3}{2}-\dfrac{5}{6}\left(\dfrac{1}{2}\right)^2+\sqrt{4}=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{1}{4}+2=\dfrac{3}{2}-\dfrac{5}{24}+2=\dfrac{36}{24}-\dfrac{5}{24}+\dfrac{48}{24}=\dfrac{79}{24}\)c)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}=1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{32}{17}=1+1-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{34}{17}-\dfrac{32}{17}=\dfrac{2}{17}\)d)\(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)=-8.\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-8.\dfrac{2}{4}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)=-8.\dfrac{2}{4}:\dfrac{13}{12}=-4.\dfrac{12}{13}=\dfrac{-48}{13}\)e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)+28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)=\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right).\left(-\dfrac{5}{3}\right)=\left(\dfrac{120}{7}+\dfrac{196}{7}\right).\left(-\dfrac{5}{3}\right)=\dfrac{316}{7}.\left(-\dfrac{5}{3}\right)=-\dfrac{1580}{21}\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
Toàn câu dễ nên bạn tự làm đi.
Trong lúc bạn đánh xong bài này thì bạn có thể làm xong rồi đó.
Đừng có ỷ lại vào người khác ,động não lên.
a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)
\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)
\(=-10\)
b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{5}-\dfrac{3}{7}-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=-\dfrac{3}{5}:\dfrac{4}{5}\)
\(=-\dfrac{3}{4}\)
c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{20}{3}\)
d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{1}{1}:\dfrac{1}{144}\)
\(=144\)
e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)
= 14 . \(\left(-\dfrac{5}{7}\right)\)
= -10
b) = (\(-\dfrac{2}{5}+-\dfrac{1}{5}\)) + \(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)\): \(\dfrac{4}{5}\)
= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)
= \(\dfrac{3}{4}\)
c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)
= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)
= \(-\dfrac{4}{3}\)
d) = 1: \(\dfrac{23}{48}\)
=\(\dfrac{48}{23}\)
e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)
= \(\left(-\dfrac{10}{3}\right)-\dfrac{7}{3}\)
=\(-\dfrac{17}{3}\)
f) = 10 485.76
e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)
=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)
=\(-\dfrac{516}{7}\)
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)
=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)
=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)
=\(\dfrac{119}{240}\)
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.
b: \(=\left(52.5-1002.25\right)+\left(48.7-82.47-8.23\right)+318\)
\(=-949,75-42+318=-673.75\)
c: \(D=\left(-1\right)+\left(-1\right)+...+\left(-1\right)=-30\)
d: \(=17-\left\{15-\dfrac{2}{3}+5+\dfrac{4}{3}-13\right\}+\dfrac{2}{3}\)
\(=17-\left\{7+\dfrac{2}{3}\right\}+\dfrac{2}{3}=10\)