Viết số thích hợp vào chỗ chấm: Hai số có tổng là 64 và tỉ số là 3/5. Hai số đó là ..............

\(A=-1-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{1024}\)

\(A=-1-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)

\(A=-1-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

Đặt:

\(B=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2B=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(2B=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2B-B=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(B=1-\dfrac{1}{2^{10}}\)

Thay B vào A

\(A=-1-\left(1-\dfrac{1}{2^{10}}\right)\)

\(A=-1-1+\dfrac{1}{2^{10}}\)

\(A=-2+\dfrac{1}{2^{10}}\)

Có cần lắm ko,nhìn bài của Nhung ảo

\(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|=0\)

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|y+\dfrac{2}{3}\right|\ge0\forall y\\\left|x^2+xz\right|\ge0\forall x;z\end{matrix}\right.\) \(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|y+\dfrac{2}{3}\right|=0\\\left|x^2+xz\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{2}{3}\\z=-\dfrac{1}{2}\end{matrix}\right.\)

\(VT=12^8.9^{12}=\left(2^2.3\right)^8.\left(3^2\right)^{12}=2^{16}.3^8.3^{24}=2^{16}.3^{32}=2^{16}.\left(3^2\right)^{16}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^6=VP\)

\(E=1.2+2.3+3.4+...+99.100\)

\(3E=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)

\(3E=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)

\(3E=99.100.101\)

\(E=\dfrac{99.100.101}{3}\)

\(E=333300\)

x = 2 

tich nha

\(M=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(M=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(M=\left[x\left(x+5\right)+2\left(x+5\right)\right]\left[x\left(x+4\right)+3\left(x+4\right)\right]-24\)

\(M=\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)-24\)

\(M=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(M=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-24\)

\(M=\left(x^2+7x+11\right)^2-1-24\)

\(M=\left(x^2+7x+11\right)^2-25\)

\(M=\left(x^2+7x+11+5\right)\left(x^2+7x+11-5\right)\)

\(M=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

Ta có  / x-2013/ + /x -2020/ = /x -2013/ + / 2020 -x/ >/   / x -2013 + 2020 -x / = 7

Dấu ' =' xảy ra khi   2013</ x </ 2020

Nếu x là số tự nhiên 

=> x thuộc { 2013; 2014;2015;....2020}