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a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
Mk thấy bài 1 và 2 dễ nên bạn tự làm nha
3
+)Ta có n-2 \(⋮\)n-2
=>2.(n-2)\(⋮\)n-2
=>2n-4\(⋮\)n-2(1)
+)Theo bài ta có:2n+1\(⋮\)n-2(2)
+)Từ (1) và (2)
=>(2n+1)-(2n-4)\(⋮\)n-2
=>2n+1-2n+4\(⋮\)n-2
=>5\(⋮\)n-2
=>n-2\(\in\)Ư(5)={\(\pm\)1;\(\pm\)5}
+)Ta có bảng:
n-2 | -1 | 1 | -5 | 5 |
n | 1\(\in\)Z | 3\(\in\)Z | -3\(\in\)Z | 7\(\in\)Z |
Vậy n\(\in\){1;3;-3;7}
Chúc bn học tốt
a. 5.(–8).( –2).(–3) b. 4.(–5)2 + 2.(–5) – 20
=(-5).8.(-2).(-3) ={(-5).2} {4+1}-20
=(-5)(-2)(-3).8 =(-10).5-20=-50-20=-70
=10.(-24)=-240
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
Bài 1:
a)
\(A=\dfrac{-5}{6}\cdot\dfrac{3}{10}\\ =\dfrac{\left(-5\right)\cdot3}{6\cdot10}\\ =\dfrac{-1}{4}\)
b)
\(B=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12}\\ =\dfrac{4}{12}-\dfrac{3}{12}+\dfrac{1}{12}\\ =\dfrac{4-3+1}{12}\\ =\dfrac{1}{6}\)
Bài 2:
\(A=\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{4}{5}-\dfrac{14}{5}\right|:\dfrac{8}{3}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{-10}{5}\right|\cdot\dfrac{3}{8}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+2\cdot\dfrac{3}{8}\\ =\dfrac{-3}{4}+\dfrac{3}{4}\\ =0\)
\(B=\left(\dfrac{-1}{2}\right)^2:1\dfrac{3}{8}+25\%\cdot\dfrac{3}{11}\\ =\left(\dfrac{-1}{2}\right)^2:\dfrac{11}{8}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{1}{4}\cdot\dfrac{8}{11}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{8}{44}+\dfrac{9}{44}\\ =\dfrac{17}{44}\)
\(C=\dfrac{-8}{5}+0,6+\left|\dfrac{-1}{2}\right|+\dfrac{1}{2}\\ =\dfrac{-8}{5}+\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(\dfrac{-8}{5}+\dfrac{3}{8}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\\ =\left(-1\right)+1\\ =0\)
\(D=\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}:\dfrac{13}{11}+1\dfrac{5}{9}\\ =\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}\cdot\dfrac{11}{13}+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot\left(\dfrac{2}{13}+\dfrac{11}{13}\right)+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot1+\dfrac{14}{9}\\ =\dfrac{-5}{9}+\dfrac{14}{9}\\ =1\)
3:
\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)
4:
=>2/3:x=-2-1/3=-7/3
=>x=-2/3:7/3=-2/7
5:
AC=CB=12/2=6cm
IB=6/2=3cm