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a)-91 phần200
b)-25phần 4
c)5 phần 2
d)2
e)0
a, ( 0,36-2,18) : ( 3,8 + 0,2)
= -1,82 : 4
=-0,455 hay -91/200
b, 3/8*19/1/3-3/8*33/1/3
=3/8*(19/1/3-33/1/3)
=3/8*(-14)
=-21/4
a/ \(\dfrac{\dfrac{-5}{12}}{\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|}=\dfrac{\dfrac{-4}{9}}{\dfrac{8}{15}}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\left(-\dfrac{5}{12}\right).\left(\dfrac{8}{15}\right)\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\dfrac{-2}{9}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\\dfrac{2}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\\dfrac{2}{3}x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy ....
2) -12:\(\left(-\dfrac{5}{6}\right)^2\)=\(-12:\dfrac{25}{36}=-12\cdot\dfrac{36}{25}=-\dfrac{432}{25}\)
s) \(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
= \(-\dfrac{1}{12}-\dfrac{55}{24}=-\dfrac{2}{24}-\dfrac{55}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)
t) \(-1,75-\left(-\dfrac{1}{9}-2\dfrac{1}{18}\right)=-1,75-\left(-\dfrac{2}{18}-\dfrac{37}{18}\right)\)
= -1,75-(\(-\dfrac{13}{6}\)) = \(-\dfrac{7}{4}+\dfrac{13}{6}=\dfrac{5}{12}\)
c) \(\left(\sqrt{\dfrac{1}{9}}-0,5\right)^3+\dfrac{-1}{3}=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^3-\dfrac{1}{3}\)
= \(\left(-\dfrac{1}{6}\right)^3-\dfrac{1}{3}=\dfrac{-1}{216}-\dfrac{1}{3}=-\dfrac{73}{216}\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)
= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)
\(\left\{{}\begin{matrix}1-\dfrac{2}{3}=\dfrac{1}{3}\\1-\dfrac{3}{4}=\dfrac{1}{4}\\1-\dfrac{4}{5}=\dfrac{1}{5}\\1-\dfrac{9}{10}=\dfrac{1}{10}\end{matrix}\right.\)
Vì:
\(\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}>...>\dfrac{1}{10}\)
nên:
\(\dfrac{2}{3}< \dfrac{3}{4}< \dfrac{4}{5}< ...< \dfrac{9}{10}\)
a)
Ta có:
\(\)\(\left\{{}\begin{matrix}\dfrac{3}{4}=\dfrac{2+1}{3+1}\\\dfrac{4}{5}=\dfrac{3+1}{4+1}\\\dfrac{5}{6}=\dfrac{4+1}{5+1}\\\dfrac{9}{10}=\dfrac{8+1}{9+1}\end{matrix}\right.\)
Suy ra quy luật:
Phân số tiếp theo chính là tử của p/s ban đầu +1/mẫu của p/s ban đầu +1
Vậy phân số sau phân số \(\dfrac{a}{b}\) là \(\dfrac{a+1}{b+1}\)
So sánh :
\(\dfrac{a}{b}\) và \(\dfrac{a+1}{b+1}\)
\(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\)
\(\dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\)
Vậy cần so sánh:
\(\dfrac{ab+a}{b^2+b}\) với \(\dfrac{ab+b}{b^2+b}\)
Cần so sánh:
\(ab+a\) và \(ab+b\)
Cần so sánh \(a\) với \(b\)
Nếu \(a>b\Rightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\)
Nếu \(a< b\Rightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)
Nếu \(a=b\) \(\Rightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}=1\)
Còn cách khác ngắn hơn nhưng lười làm lắm :v
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
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) \(\dfrac{12}{\left(-2\right)^n}=\dfrac{-12}{8}\)
\(\Rightarrow12.8=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow96=\left(-2\right)^n.\left(-12\right)\)
\(\Rightarrow\left(-2\right)^n=\dfrac{96}{-12}\)
\(\Rightarrow\left(-2\right)^n=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
2)
a) \(\dfrac{4}{9}\) và \(\dfrac{5}{8}\) Mẫu chung: 72
\(\dfrac{4}{9}=\dfrac{4.8}{72}=\dfrac{32}{72}\)
\(\dfrac{5}{8}=\dfrac{5.9}{72}=\dfrac{45}{72}\)
Vì \(\dfrac{32}{72}< \dfrac{45}{72}\)
Vậy \(\dfrac{4}{9}< \dfrac{5}{8}\)
b) \(-\sqrt{\dfrac{4}{9}}\) và \(\dfrac{-3}{4}\) MTC: 12
\(-\sqrt{\dfrac{4}{9}}=-\sqrt{\left(\dfrac{2}{3}\right)^2}=-\dfrac{2}{3}=\dfrac{-2.4}{12}=\dfrac{-8}{12}\)
\(-\dfrac{3}{4}=\dfrac{-3.3}{12}=\dfrac{-9}{12}\)
Vì \(\dfrac{-8}{12}>\dfrac{-9}{12}\)
Vậy \(-\sqrt{\dfrac{4}{9}}>\dfrac{-3}{4}\)