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Bài 1:
c) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.4^4.5^4}{5^{10}.4^5}=\dfrac{5^8.4^4}{5^8.5^2.4^4.4}=\dfrac{1}{25.4}=\dfrac{1}{100}\)
Bài 2: a) \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\\\left(y+0,4\right)^{100}\ge0\forall y\\\left(z-3\right)^{678}\ge0\forall z\end{matrix}\right.\) \(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)
Vậy ...
Bài 3: \(M=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
\(=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256.\)
Vậy \(M=256.\)
Mấy bài kia dễ tự làm.
\(3)\)
\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256\)\(4)\)
\(2^{24}=\left(2^6\right)^4=64^4;3^{16}=\left(3^4\right)^4=81^4\)
\(\Leftrightarrow2^{24}< 3^{16}\)
\(\left\{{}\begin{matrix}1-\dfrac{2}{3}=\dfrac{1}{3}\\1-\dfrac{3}{4}=\dfrac{1}{4}\\1-\dfrac{4}{5}=\dfrac{1}{5}\\1-\dfrac{9}{10}=\dfrac{1}{10}\end{matrix}\right.\)
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
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!!!
a/ Ta có :
\(\dfrac{1}{2017}>0\)
\(-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{1}{2017}>\dfrac{-1}{2}\)
b/ \(\dfrac{499}{-497}< -1\)
\(\dfrac{-23456}{23456}=-1\)
\(\Leftrightarrow\dfrac{-497}{499}< \dfrac{-23456}{23456}\)
a, Ta có \(\dfrac{1}{2017}>0\)
\(-\dfrac{1}{2}< 0\)
<=>\(\dfrac{1}{2017}>\left(\dfrac{-1}{2}\right)\)
b,\(\dfrac{499}{-497}>\left(-1\right)\)
=>\(\dfrac{-23456}{23456}=-1\)
<=>\(\dfrac{-497}{499}< \dfrac{-23456}{23456}\)
Chúc Bạn Học Tốt!!!
Bài 1:
a) \(\mathbb{Z}\)
b) \(\mathbb{Q}\)
c) \(\mathbb{Q}\)
d) \(\mathbb{Z}\)
e) \(\mathbb{N}\)
Bài 2:
a) Ta thấy: \(\frac{1}{8}>0; \frac{-3}{8}< 0\) \(\Rightarrow \frac{1}{8}> \frac{-3}{8}\)
b) \(\frac{-3}{7}< 0; 2\frac{1}{2}>0\Rightarrow \frac{-3}{7}< 2\frac{1}{2}\)
c) \(-3,9< 0< 0,1\)
d) \(-2,3< 0< 3,2\)
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 ...