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1/ a/ \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...........
b/ \(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\left(\dfrac{2}{3}x-\dfrac{1}{5}\right)^5=\left(\dfrac{1}{3}\right)^5\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{24}{30}\)
Vậy ....
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
555 ^ 2 ≡ 5 (mod 10)
555 ^3≡5 (mod 10)
555^5=555^2.555^3≡5.5≡5 (mod 10)
~~> 555^777≡5 (mod 10)
Suy ra
333^555^777đồng dư với 333^5
Do 333^5=333^2.333^3≡3 (mod10)
Vậy chữ số tận của 333^555^777 là 3 . (1)
Làm tương tự ta được 777^555^333 có chữ số tận cùng là 7 (2)
(1) và (2) Suy ra 333^555^777 +777^555^333 có chữ số tận cùng là 0
Vậy 333^555^777 +777^555^333 chia hết cho 10.
\(555\equiv-1\left(\text{mod 4}\right)\Rightarrow555^{777}\equiv\left(-1\right)^{777}\left(\text{mod 4}\right)\equiv\left(-1\right)\left(\text{mod 4}\right)\)
\(\Rightarrow\text{555^777 chia 4 dư 3. }\)
\(555^{333}\equiv\left(-1\right)^{333}\left(\text{mod 4}\right)\equiv\left(-1\right)\left(\text{mod 4}\right)\)
\(\Rightarrow\text{555^333 chia 4 dư 3}\)
\(\text{Đến đây dễ rồi -__-}\)
Ta có:
5552≡5 (mod 10)
5553≡5( mod 10)
5555=5552.5553≡5.5≡5(mod 10)
---> 555777≡5(mod 10)
Suy ra:
333555777đồng dư với 3335
Do 3335=3332.3333≡3(mod 10)
Vậy chữ số tận cùng của 333555777là 3 (1)
Làm tương tự với 777555333có chữ số tận cùng là 7 (2)
Từ (1) và (2) suy ra 333555777+777555333có chữ số tận cùng là 0
Vậy 333555777+777555333chia hết cho 10 (đpcm)
\(a)\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{12}{24}+\left(-\dfrac{5}{24}\right)\)
\(=\dfrac{18+12+\left(-5\right)}{24}\)
\(=\dfrac{25}{24}\)
\(b)\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}.\dfrac{-2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\left(\dfrac{-2}{13}+\dfrac{-11}{13}+\dfrac{13}{13}\right)\)
\(=\dfrac{5}{7}.0=0\)
\(c)\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
\(d)\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{38}{51}+\dfrac{306}{714}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{532}{714}+\dfrac{306}{714}\)
\(=\dfrac{1391}{714}\)
a)\(\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}=\dfrac{18}{24}+\dfrac{12}{24}-\dfrac{5}{24}=\dfrac{25}{24}\)
b)\(\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}=\dfrac{5}{7}\left(\dfrac{-2}{13}-\dfrac{11}{13}+1\right)=\dfrac{5}{7}.0=0\)
c)\(\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=2,5\)
d)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}=\dfrac{15}{34}+\left(\dfrac{1}{3}+\dfrac{38}{51}+\dfrac{3}{7}\right)=\dfrac{15}{34}+\dfrac{538}{357}=\dfrac{1391}{714}\)
1) \(25.\left(-\dfrac{1}{5}\right)^3+\dfrac{1}{5}-2.\left(-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\)
\(=25.\left(-\dfrac{1}{125}\right)+\dfrac{1}{5}-2.\left(\dfrac{1}{4}\right)-\dfrac{1}{2}\)
\(=-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{2}-\dfrac{1}{2}\)
\(=-1\)
2) \(\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{8}:\left(\dfrac{4}{18}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{15}{36}\right)\)
\(=\dfrac{7}{8}:\dfrac{3}{18}+\dfrac{7}{8}:\left(-\dfrac{7}{18}\right)\)
\(=\dfrac{7}{8}:\left(\dfrac{3}{18}-\dfrac{7}{18}\right)\)
\(=\dfrac{7}{8}:\left(-\dfrac{2}{9}\right)\)
\(=\dfrac{7}{8}.\left(-\dfrac{9}{2}\right)\)
\(=-\dfrac{63}{16}\)
Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 .
= ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) .
= 2 + ( 20/21 + 7/21 ) .
= 2 + 9/7 .
= 23/7 .
b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) .
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41 .
= 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) .
= 1/3 - 1 + 1 .
= 1/3 .
c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 .
= [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 .
= [ 11/4 . ( - 1 ) ] . 8/33 .
= -11/4 . 8/33 .
= -2/3 .
d, 38/45 - ( 8/45 - 17/51 - 3/11 ) .
= 38/45 - 8/45 + 17/51 + 3/11 .
= 2/3 + 17/51 + 3/11 .
= 374/561 + 187/561 + 153/561 .
= 14/11 .
1,\(\dfrac{a}{b}=\dfrac{x}{y}\) khi ay=bx
2,
a,x=\(\dfrac{-1.12}{4}\)
x=\(\dfrac{-12}{4}=-3\)
b,\(\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^5\)
\(\Rightarrow\)2x-1=5
2x=6
x=6:2=3
c,\(\dfrac{4}{7}\).x=\(\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
x=\(\dfrac{13}{15}.\dfrac{7}{4}=\dfrac{91}{60}\)
3,ta có:\(5^{202}=\left(5^2\right)^{101}\)=\(25^{101}\)
2\(^{505}\)=\(\left(2^5\right)^{101}\)=\(32^{101}\)
vì 25<32 nên \(25^{101}< 32^{101}\) hay \(5^{202}< 2^{505}\)
1) \(\dfrac{a}{b}=\dfrac{x}{y}\) khi \(a.y=b.x\)
2) \(a,\dfrac{x}{12}=\dfrac{-1}{4}\)
\(\Rightarrow4x=-12\)
\(\Rightarrow x=-\dfrac{12}{4}=-3\)
Vậy x = -3
\(b,\left(\dfrac{1}{3}\right)^{2x-1}=\dfrac{1}{243}\)
\(\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^5\)
\(\Rightarrow2x-1=5\)
\(\Rightarrow x=\dfrac{5-1}{2}=2\)
Vậy x = 2
\(c,\dfrac{4}{7}x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}x=\dfrac{13}{15}\)
\(\Rightarrow x=\dfrac{13}{15}:\dfrac{4}{7}=1\dfrac{31}{60}\)
Vậy \(x=1\dfrac{31}{60}\)
3) So sánh \(5^{202}\) và \(2^{505}\)
\(5^{202}=\left(5^2\right)^{101}=25^{101}\)
\(2^{505}=\left(2^5\right)^{101}=32^{101}\)
\(\Rightarrow25^{101}< 32^{101}\)
\(\Rightarrow5^{202}< 2^{505}\)
=5/7+3/5+1/5=5/7+4/5=25/35+28/35=53/35