2, 100^2+200^2+300^2+..+1000^2

=100^2+2^2×100^2+3^2×100^2+...+100^2×10^2

=100^2×( 1^2+2^2+3^2+..+10^2)

=100^2×385

= 3850000

1)

a. \(\left(3x^2-50\right)^2=5^4\)

\(\Leftrightarrow3x^4-50=625\)

\(\Leftrightarrow3x^4=675\)

\(\Leftrightarrow x^4=225\)

\(\Leftrightarrow x=\sqrt{15}\) 

2)

a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)

\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)

b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)

\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)

a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)

vậy...

2.

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

⇒ \(5x+1=\pm\frac{6}{7}\)

⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)

Chúc bạn học tốt!

a) \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)

\(\Rightarrow\frac{1}{3^m}=\frac{1}{81}\)

<=> 3^m = 81

=> 3^m = 3⁴ ( 81 = 3⁴ )

<=> m = 4

b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

\(\left(\frac{3}{5}\right)^n=\frac{9}{9765625}\)

\(\Rightarrow\frac{3}{5^n}=\frac{9}{9765625}\)

=> 5ⁿ = 9765625

=> 5ⁿ = 5¹⁰ ( 9765625 = 5¹⁰ )

<=> n = 10

\(\left(-0,25\right)^p=\frac{1}{256}\)

\(\left(\frac{-1}{4}\right)^p=\frac{1}{256}\)

\(\Rightarrow\frac{-1}{4^p}=\frac{1}{256}\)

=> 4^p = 256

=> 4^p = 4⁴ ( 256 = 4⁴ )

<=> p = 4