\(\left(2x+1\right)^2+\left(b+3\right)^4=0\)

Mà \(\left(2a+1\right)^2\ge0\forall x;\left(b+3\right)^4\ge0\forall b\)

\(\left(2a+1\right)^2+\left(b+3\right)^4=0\)chỉ khi: \(\hept{\begin{cases}\left(2a+1\right)^2=0\Rightarrow2a+1=0\Rightarrow a=\frac{-1}{2}\\\left(b+3\right)^4=0\Rightarrow b+3=0\Rightarrow b=-3\end{cases}}\)

\(\left(a-7\right)^2+\left(3b+2\right)^2+\left(4c-5\right)^6\le0\)

Xét: \(\left(a-7\right)^2+\left(3b+2\right)^2+\left(4c-5\right)^6< 0\)=> Vô lý

Xét: \(\left(a-7\right)^2+\left(3b+2\right)^2+\left(4c-5\right)^6=0\)

\(\Rightarrow\left(a-7\right)^2=0\Rightarrow a-7=0\Rightarrow a=7\)

\(\Rightarrow\left(3b+2\right)^2=0\Rightarrow3b+2=0\Rightarrow3b=-2\Rightarrow b=\frac{-2}{3}\)

\(\Rightarrow\left(4c-5\right)^6=0\Rightarrow4c-5=0\Rightarrow4c=5\Rightarrow c=\frac{5}{4}\)

11⁶ - 11⁵ + 11⁴

= 11⁴ . 11² - 11⁴ . 11 + 11⁴ . 1

= 11⁴ . 121 - 11⁴ . 11 + 11⁴ . 1

= 11⁴ . ( 121 - 11 + 1 )

= 11⁴ . 111

Ta thấy : 111 \(⋮\)111

\(\Rightarrow\)11⁴ . 111  \(⋮\)111 hay 11⁶ - 11⁵ + 11⁴ \(⋮\)111

nhanh nha 3 h mik cần r

ta có :

\(|x-\frac{2}{3}|+|x+y+\frac{3}{4}|+|y-z-\frac{5}{6}|=0\Leftrightarrow\hept{\begin{cases}x-\frac{2}{3}=0\\x+y+\frac{3}{4}=0\\y-z-\frac{5}{6}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=-\frac{17}{12}\\z=-\frac{9}{4}\end{cases}}\)

giúp mình với ạ

a) 3x = 4 + 2

3x = 6

=> x = 2;