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Giải:
Bài 1:
\(707-3\left(x+3\right)=407\)
\(\Leftrightarrow3\left(x-3\right)=300\)
\(\Leftrightarrow x+3=100\)
\(\Leftrightarrow x=97\)
Vậy \(x=97\).
Bài 2:
a) \(143-43.\left[\left(25:5\right)^2-5^2\right]\)
\(=143-43.\left(5^2-5^2\right)\)
\(=143-43.1\)
\(=143-43=100\)
b) \(7:7+49\left(2^3.15-5.4\right)\)
\(=1+49.\left(120-20\right)\)
\(=1+49.100\)
\(=4901\)
c) \(\left(2^3.9^4+9^3.45\right):\left(9^2.10-9^2\right)\)
\(=\left(8.9^4+9^4.5\right):\left(9^2.10-9^2\right)\)
\(=\left[9^4\left(8+5\right)\right]:\left[9^2\left(10-1\right)\right]\)
\(=9^4.13:9^3\)
\(=9.13=117\)
Chúc bạn học tốt!
e) \(5^{13}:5^{10}-25.2^2\)
\(=5^3-5^2.2^2\)
\(=5^2\left(5-2^2\right)\)
\(=25\)
e) 513 : 510 - 25 . 22 f) 20:22+59:58 g) 100+ 52+ 7.32 h) 84:4+39:37+50 i) 29- [16+ 3. (51- 49)]
= 53 - 25 . 4 = 20 :4 + 5 = 100+ 25+ 7.9 = 21+ 32+ 1 = 29- [16+ 3.2 ]
=125 - 100 = 5+ 5 = 125+ 63 = 21+ 9+ 1 = 29- [16+ 6]
=25 = 10 = 188 = 31 = 29 - 22
=7
\(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
\(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{5^4\cdot4^4\cdot5^4}{5^5\cdot5^5.4^5}=\frac{1}{100}\)
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2=3-1+\frac{1}{8}=\frac{17}{8}\)
\(\left(\frac{3}{9}.x-5\right)^2=\frac{9}{49}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{9}x-5=\frac{3}{7}\\\frac{3}{9}x-5=-\frac{3}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{9}x=\frac{3}{7}+5\\\frac{3}{9}x=-\frac{3}{7}+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{9}x=\frac{38}{7}\\\frac{3}{9}x=\frac{32}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{114}{7}\\x=\frac{96}{7}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=\frac{114}{7}\\x=\frac{96}{7}\end{cases}}\)
Chúc bạn học tốt !!!!