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Bài 1:

\(\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)

\(=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)

\(=\dfrac{2^7+12^3}{2^7\cdot5^2+512}\)

\(=\dfrac{1856}{3712}\)

\(=0,5\)

Bài 2:

\(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Rightarrow5x+1=\dfrac{6}{7}\)

\(\Rightarrow5x=\dfrac{-1}{7}\)

\(\Rightarrow x=\dfrac{-1}{35}\)

13 tháng 9 2021

\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

13 tháng 9 2021

\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)

\(A=\dfrac{\left(\dfrac{2}{5}\right)^7.5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7.5^2+512}\\ A=\dfrac{128+1728}{153+512}\\ A=\dfrac{1856}{665}=2\dfrac{526}{665}\)

Chúc bạn học tốt!!!hihi

7 tháng 8 2017

\(\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7+5^2+512}\)

Sửa đề: (2/7)^7*7^7

\(A=\dfrac{\left(2\right)^7+\left(\dfrac{9}{3}:\dfrac{3}{16}\right)^3}{2^7\left(5^2+2^2\right)}\)

\(=\dfrac{\left(2\right)^7+\left(16\right)^3}{2^7\cdot29}\)

\(=\dfrac{2^7+2^7\cdot2^5}{2^7\cdot29}=\dfrac{1+2^5}{29}=\dfrac{33}{29}\)

14 tháng 8 2017

\(A=\dfrac{\left(\dfrac{2}{5}\right)^7.5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7.5^2+512}\)

\(A=\dfrac{128+1728}{153+512}\)

\(A=\dfrac{1856}{665}\)

\(A=2\dfrac{526}{665}\)

9 tháng 9 2017

Ta có:

\(\dfrac{\left(\dfrac{2}{5}\right)^7.5^7+\left(\dfrac{9}{4}\right)^3\div\left(\dfrac{3}{16}\right)^3}{2^7.5^7+512}\)

=\(\dfrac{\left(\dfrac{2}{5}.5\right)^7+\left(\dfrac{9.16}{4.3}\right)^3}{2^7.5^7+2^9}\)=\(\dfrac{2^7+12^3}{2^7.5^7+2^9}\)=\(\dfrac{2^7+2^6.3^3}{2^7.5^7+2^9}\)

=\(\dfrac{2^6.\left(2+3^3\right)}{2^6.\left(2.5^7+2^3\right)}\)=\(\dfrac{29}{156258}\).

Hì hì sai ko bít nha

18 tháng 7 2017

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

~ Học tốt ~

18 tháng 7 2017

Bài 1:

1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)

\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)

\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)

\(=3^2=9\)

2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)

\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)

\(=2^7:2^3:\dfrac{1}{2^4}\)

\(=2^4.2^4=256\)

3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)

\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)

\(=\dfrac{43}{48}\)

4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)

\(=-3-1+\dfrac{1}{8}\)

\(=-4+\dfrac{1}{8}\\ \)

\(=-\dfrac{31}{8}\)

5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)

Chúc bạn học tốt haha

\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)

29 tháng 10 2021

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

28 tháng 9 2021

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)