Đỗ Bình An
Giới thiệu về bản thân
102 + 2(x - 1) = 2^2 x 50
=> 102 + 2(x - 1) = 4 x 50
=> 102 + 2(x - 1) = 200
=> 2(x - 1) = 200 - 102
=> 2(x - 1) = 198
=> x - 1 = 198:2
=> x - 1 = 99
=> x = 99 + 1
=> x =100
Vậy..
\(\dfrac{3}{4}\times\dfrac{24}{12}+\dfrac{4}{5}:\dfrac{12}{25}-\dfrac{1}{2}\)
\(=\dfrac{3}{4}\times\dfrac{24}{12}+\dfrac{4}{5}\times\dfrac{25}{12}-\dfrac{1}{2}\)
\(=\dfrac{1}{1}\times\dfrac{3}{2}+\dfrac{1}{1}\times\dfrac{5}{3}-\dfrac{1}{2}\)
\(=\dfrac{3}{2}+\dfrac{5}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{3}{2}-\dfrac{1}{2}\right)+\dfrac{5}{3}\)
\(=1+\dfrac{5}{3}\)
\(=\dfrac{8}{3}\)
\(\dfrac{27^4\times15^3\times8^2}{6^7\times9^3\times15}\)
\(=\dfrac{\left(3^3\right)^4\times15^3\times\left(2^3\right)^2}{\left(2\times3\right)^7\times\left(3^2\right)^3\times15}\)
\(=\dfrac{3^{12}\times15^3\times2^6}{2^7\times3^7\times3^6\times15}\)
\(=\dfrac{3^5\times15^2\times1}{2\times1\times3^6\times1}\)
\(=\dfrac{1\times15^2\times1}{2\times1\times3\times1}\)
\(=\dfrac{225}{6}\)
(P/S: Đây là lần đầu mik làm nên có 1 chút sai sót ạ)
\(4x^2-36=0\)
\(4x^2\) \(=0+36\)
\(4x^2\) \(=36\)
\(x^2\) \(=36:4\)
\(x^2\) \(=9\)
\(x^2\) \(=3^2\)
\(=>x=3\)
Vậy...
\(\dfrac{1}{2}-\dfrac{43}{101}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}+\left(-\dfrac{43}{101}\right)\)
\(=0+\left(-\dfrac{43}{101}\right)\)
\(=-\dfrac{43}{101}\)