\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(25.4\right)^2}{\left(5.\left(-2\right)\right)^5}=\dfrac{\left(100\right)^2}{\left(-10\right)^5}=\dfrac{1}{-10}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
1.
$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$
$=\frac{1}{4^{10}}$
2.
$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$
3.
$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)
c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)
\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)
\(=\dfrac{125}{70}=\dfrac{25}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa:\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow\left(x-5\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)
\(=1-\dfrac{-5}{2}\)
\(=\dfrac{2}{2}-\dfrac{-5}{2}\)
\(=\dfrac{7}{2}\)
dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!
Giải:
a)(4-12/5).25/8-2/5:-4/25
=8/5.25/8-(-5/2)
=5+5/2
=15/2
b)(-5/24+3/4-7/12):(-5/16)
=-1/24:(-5/16)
=2/15
c)6/7+5/4:(-5)-(-1/28).(-2)2
=6/7+(-1/4)-(-1/28).4
=6/7-1/4-(-1/7)
=6/7-1/4+1/7
=(6/7+1/7)-1/4
=1-1/4
=3/4
Chúc bạn học tốt!
ủa thế đề là j hả bn