Ta có |5x-5012| = |-5x+5012| >_ -5x +5012.Dấu "=" khi -5x+5012>0

         | 5x+300|>_ 5x+300.Dấu "=" khi 5x+300>0

=> |-5x+5012| + |5x+300| >_ -5x+5012 +5x + 300

=> A >_ 5312

Dấu "=" khi -5x+5012>0 => x<5012/5

                   5x+300> 0   => x>-60

Vậy Min A = 5312 khi -60<x<5012/5

\(\left(0,75\right)^3\cdot1024\)

\(=\frac{27}{64}\cdot1024\)

\(=432\)

0,75 ^ 3 x 1024

= 0,421875 x 1024

= 432

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{5}\)

\(=>\frac{x.y}{2.5}=\frac{10}{10}=1\)

Ta có :\(x=1.2=2\)

           \(y=1.5=5\)

Vậy \(x=2;y=5\)

Ai k sai giải thích với ạ :(

Ko thì đừng có k bừa bãi

#SMC

\(\left(x+1\right)^2=\left(x+1\right)^4\)

\(\Rightarrow\left(x+1\right)^4-\left(x+1\right)^2=0\)

\(\Rightarrow\left(x+1\right)^2\left[\left(x+1\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x+1\right)^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x+1=\pm1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=0\text{ or }x=-2\end{cases}}\)

Đặt \(A=1^2+2^2+3^2+...+10^2=\left[1\cdot1\right]^2+\left[2\cdot1\right]^2+\left[3\cdot1\right]^2+...+\left[10\cdot1\right]^2\)

\(=1^2\cdot1^2+2^2\cdot1^2+3^2\cdot1^2+...+10^2\cdot1^2\)

\(=1^2\left[1^2+2^2+3^2+...+10^2\right]=1\cdot385=385\)

\(S=2^2+4^2+6^2+...+20^2\)

\(=\left[2\cdot1\right]^2+\left[2\cdot2\right]^2+\left[2\cdot3\right]^2+...+\left[2\cdot10\right]^2\)

\(=2^2\cdot1^2+2^2\cdot2^2+2^2\cdot3^2+...+2^2\cdot10^2\)

\(=2^2\left[1^2+2^2+3^2+...+10^2\right]=4\cdot385=1540\)