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\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
Vậy \(P=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)
\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)
\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)
\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)
Vậy \(Q=\dfrac{29}{125}\)
1.
\(\left(1-2x\right)^4=81\\ \Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-2x=2\\-2x=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy ...
2.
\(\left|2,5-x\right|=1,3\\ \Rightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)
Vậy ...
3.
\(5^x+5^{x+2}=650\\ 5^x\left(1+5^2\right)=650\\ 5^x\cdot26=650\\ 5^x=25\\ x=2\)
Vậy ...
4, 5 tự làm
b: \(\left|x-\dfrac{3}{5}\right|< \dfrac{1}{3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{5}>-\dfrac{1}{3}\\x-\dfrac{3}{5}< \dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\dfrac{4}{15}< x< \dfrac{14}{15}\)
c: \(\left|x+\dfrac{11}{2}\right|>-5.5\)
mà \(\left|x+\dfrac{11}{2}\right|\ge0\forall x\)
nên \(x\in R\)
Bài 1:
a) Ta có:
\(-3\dfrac{1}{5}=-2,8\)
\(\dfrac{37}{10}=3,7\)
Vì \(-3,25< -2,8< 3,7\)
Nên \(-3,25< -3\dfrac{1}{5}< \dfrac{37}{10}\)
b) Ta có:
\(\dfrac{-567}{568}>-1\)
\(\dfrac{-568}{567}< -1\)
\(\Leftrightarrow\dfrac{-567}{568}>-1>\dfrac{-568}{567}\)
\(\Leftrightarrow\dfrac{-567}{568}>\dfrac{-568}{567}\)
c) \(-0,625=-\dfrac{5}{8}\)
Vì \(8>7\)
\(\Leftrightarrow\dfrac{1}{8}< \dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{-5}{8}>\dfrac{-5}{7}\)
\(\Leftrightarrow-0,625>\dfrac{-5}{7}\)
Vậy ...
Giải:
a) \(3,6-\dfrac{-5}{6}.\left(-2,4\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{-5}{6}.\left(-\dfrac{12}{5}\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\)
\(=\dfrac{12}{5}\)
Vậy ...
b) \(\dfrac{1}{4}-0,5-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=\dfrac{1}{4}-\dfrac{1}{2}-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=-\dfrac{49}{8}\)
Vậy ...
c) \(1,1-\left(-1,2\right)-\left|-1,3\right|-2\dfrac{3}{4}\)
\(=1,1+1,2-1,3-2,75\)
\(=-\dfrac{7}{4}=-1,75\)
Vậy ...
a) \(\dfrac{1}{4}+\left(-\dfrac{3}{8}\right)=\dfrac{2}{8}+\left(-\dfrac{3}{8}\right)=-\dfrac{1}{8}\)
b)\(\dfrac{2^{10}.3^4}{3^5.2^8}=\dfrac{2^8.2^2.3^4}{3^4.3.2^8}=\dfrac{2^2}{3}=\dfrac{4}{3}\)
c) \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{9}}=\dfrac{5}{10}.10-\dfrac{1}{3}=5-\dfrac{1}{3}=\dfrac{15}{3}-\dfrac{1}{3}=\dfrac{14}{3}\)
d) \(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+5\dfrac{4}{9}:\left(-\dfrac{5}{7}\right)\)
\(=\dfrac{41}{9}:\left(-\dfrac{5}{7}\right)
+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
\(=\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
\(=\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
\(=10:\left(-\dfrac{5}{7}\right)\)
\(=10.\left(-\dfrac{7}{5}\right)\)
\(=-14\)
a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}.\left(-14\right)=-6\)
c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)
\(=-10:\left(-\dfrac{5}{7}\right)\)
\(=14\)
d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)
\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)
\(=0:\dfrac{4}{5}\)
\(=0\)
a,
\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1-0,5=1,5\)
b,
\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)
c,
\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)
d,
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)
\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)
a, \(\dfrac{3}{5}-4.\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{3}\)
\(\Rightarrow4\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{4}{15}\)
\(\Rightarrow\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{3}{4}x\in\left\{-\dfrac{1}{15};\dfrac{1}{15}\right\}\)
\(\Rightarrow\dfrac{3}{4}x\in\left\{\dfrac{4}{15};\dfrac{2}{15}\right\}\Rightarrow x\in\left\{\dfrac{16}{45};\dfrac{8}{45}\right\}\)
b, \(\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{8}-\dfrac{1}{9}\)
(do \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\) với mọi \(a\in N\)*)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{9}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{2}{9}\Rightarrow2\dfrac{2}{9}-x\in\left\{-\dfrac{2}{9};\dfrac{2}{9}\right\}\)
\(\Rightarrow x\in\left\{\dfrac{22}{9};2\right\}\)
c,\(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\Rightarrow\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(\Rightarrow\dfrac{11}{15}x=\dfrac{2}{5}\Rightarrow x=\dfrac{6}{11}\)
d, \(60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\Rightarrow\dfrac{3}{5}x+\dfrac{2}{3}x=\dfrac{1}{3}.\dfrac{19}{3}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{19}{9}\Rightarrow x=\dfrac{5}{3}\)
Chúc bạn học tốt!!!
10:15 ; \(\dfrac{16}{9}\):\(\dfrac{16}{24}\) ; \(\dfrac{2}{3}\):\(\dfrac{1}{4}\) ; 16:(-4) ; 14:21 ; -5:15 ; 12:(-3) ; -1,2:3,6
10:15=\(\dfrac{2}{3}\) ;\(\dfrac{16}{24}\)=\(\dfrac{2}{3}\) ;16:(-4)=-4 ;14:21=\(\dfrac{2}{3}\) :-5:15=\(\dfrac{-1}{3}\) ;12:(-3)=-4
-1,2:3,6=\(\dfrac{-1}{3}\)
Ta có các tỉ lệ thức: \(\dfrac{10}{15}\)=\(\dfrac{16}{24}\)=\(\dfrac{14}{21}\)=\(\dfrac{2}{3}\) ;\(\dfrac{16}{-4}\)=\(\dfrac{12}{-3}\)=-4 ;\(\dfrac{-5}{15}\)=\(\dfrac{-1,2}{3,6}\)=\(\dfrac{-1}{3}\)
giá trị tuyệt đối của 5 là 5
giá trị tuyệt đối của 3.6 là 3.6
giá trị tuyệt đối của 1/7 là 1/7
giá trị tuyệt đối của -10 là 10
giá trị tuyệt đối của 0.5 là 0.5
giá trị tuyệt đối của 4 3/5 là 23
\(5=\left|5\right|\)
\(3=\left|3\right|\)
\(6=\left|6\right|\)
\(\dfrac{1}{7}=\left|\dfrac{1}{7}\right|\)
\(-10=\left|10\right|\)
\(0,5=\left|0,5\right|\)
\(4\dfrac{3}{5}=\left|4\dfrac{3}{5}\right|\)