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\(10^6-5^7=\left(2.5\right)^6-5^7=2^6.5^6-5^7=5^6\left(2^6-5\right)\)
\(=5^6\left(64-5\right)=5^6.59\)chia hết cho 59(đpcm)
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c: \(\left(-\dfrac{2}{3}\right)^3-\dfrac{1}{9}\cdot\dfrac{12}{-18}-4\cdot\dfrac{-3}{24}\)
\(=-\dfrac{8}{27}+\dfrac{1}{9}\cdot\dfrac{2}{3}+4\cdot\dfrac{3}{24}\)
\(=-\dfrac{8}{27}+\dfrac{2}{27}+4\cdot\dfrac{1}{8}\)
\(=-\dfrac{6}{27}+\dfrac{1}{2}=\dfrac{-2}{9}+\dfrac{1}{2}=\dfrac{-4+9}{18}=\dfrac{5}{18}\)
d: \(\left(-\dfrac{4}{5}\right)^2-\dfrac{-19}{45}\cdot\dfrac{-18}{38}-\left(-\dfrac{2}{3}\right)^2\)
\(=\dfrac{16}{25}-\dfrac{19}{45}\cdot\dfrac{18}{38}-\dfrac{4}{9}\)
\(=\dfrac{16}{25}-\dfrac{4}{9}-\dfrac{19}{38}\cdot\dfrac{18}{45}\)
\(=\dfrac{144-100}{225}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{44}{225}-\dfrac{1}{5}=\dfrac{44-45}{225}=-\dfrac{1}{225}\)
e: \(\dfrac{-1}{4}\cdot\dfrac{-2}{5}-3\cdot\left(-\dfrac{2}{5}\right)^2-\left(-\dfrac{1}{3}\right)^3\)
\(=\dfrac{1}{2\cdot5}-3\cdot\dfrac{4}{25}+\dfrac{1}{27}\)
\(=\dfrac{1}{10}-\dfrac{12}{25}+\dfrac{1}{27}=\dfrac{-463}{1350}\)
g: \(\dfrac{2}{-4^2}-\dfrac{14}{19}\cdot\dfrac{-38}{7}-3\cdot\left(-\dfrac{2}{5}\right)^2\)
\(=\dfrac{-2}{16}+\dfrac{14}{7}\cdot\dfrac{38}{19}-3\cdot\dfrac{4}{25}\)
\(=-\dfrac{1}{8}+4-\dfrac{12}{25}=\dfrac{679}{200}\)
h: \(\dfrac{-3}{4}\cdot\dfrac{-8}{27}-\left(-\dfrac{1}{2}\right)\cdot\left(-\dfrac{4}{5}\right)-\left(-2\right)^3\)
\(=\dfrac{3}{27}\cdot\dfrac{8}{4}-\dfrac{1}{2}\cdot\dfrac{4}{5}+8\)
\(=\dfrac{2}{9}-\dfrac{2}{5}+8=\dfrac{352}{45}\)
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Gọi số cần tìm là \(a,b\)
\(\dfrac{a}{b}=\dfrac{1}{4}\) và \(a.b=16\)Ta có: \(a=\dfrac{16}{b}\)⇒\(a=\dfrac{16}{\dfrac{b}{b}}\)=\(\dfrac{1}{4}\)⇒\(a=\dfrac{16}{b^2}=\dfrac{1}{4}\)⇒\(b^2=64\)⇒\(b=8\) Ta có: \(a.8=16\)\(a=16:8\)\(a=2\)
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b: \(X=\overline{abcd}=100\overline{ab}+\overline{cd}\)
X chia hết cho 99
=>\(100\overline{ab}+\overline{cd}⋮99\)
=>\(99\overline{ab}+\overline{ab}+\overline{cd}⋮99\)
mà \(99\overline{ab}⋮99\)
nên \(\overline{ab}+\overline{cd}⋮99\)
chứng minh trường hợp ngược lại:
\(\overline{ab}+\overline{cd}⋮99\)
\(\Leftrightarrow\overline{ab}+\overline{cd}+99\overline{ab}⋮99\)
=>\(100\overline{ab}+\overline{cd}⋮99\)
=>\(\overline{abcd}⋮99\)
a:
Đặt \(A=\overline{123x43y}\)
A chia hết cho 5 nên y=0 hoặc y=5
TH1: y=0
A chia hết cho 3
=>\(1+2+3+x+4+3+y⋮3\)
=>\(x+13⋮3\)
=>\(x\in\left\{2;5;8\right\}\)
TH2: y=5
A chia hết cho 3
=>\(x+y+13⋮3\)
=>\(x+5+13⋮3\)
=>\(x+18⋮3\)
=>\(x\in\left\{0;3;6;9\right\}\)
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A. Ta có :
8.|x|=8
=>|x| =8:8
=>|x|=1
=>x=1 hoặc x= -1
Vậy x=1 hoặc x= -1
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em ơi chưa có bài em nhé, em chưa tải bài lên lám sao mình giúp được