Answer:

\(\left|x\right|=2\Rightarrow x=\pm2\)

Với x = 2 thì: \(2^2-3.2+2=0\)

Với x = -2 thì: \(\left(-2\right)^2-3.\left(-2\right)+2=12\)

dễ thấy vế trái luôn>0 nên 6x>0=> x>0

x>0, bỏ dấu trị tuyệt đối ra ta đc 4x+10=6x

x=5

chúc bạn học giỏi, ăn Tết đc ngon, hehe -_-

HYC-30/1/2022

Answer:

\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=6x\)

Có \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\)

\(\Rightarrow6x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+1+x+2+x+3+x+4=6x\)

\(\Rightarrow4x+10=6x\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

Answer:

\(A=2\left|3x-2\right|-1\)

Có: \(\left|3x-2\right|\ge0\forall x\)

\(\Rightarrow2\left|3x-2\right|\ge0\forall x\)

\(\Rightarrow2\left|3x-2\right|-1\le-1\forall x\)

Dấu "=" xảy ra khi: \(3x-2=0\Rightarrow3x=2\Rightarrow x=\frac{2}{3}\)

\(C=\left(x-1\right)^2+\left(y+2\right)^2\)

Có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x;y}\)

\(\Rightarrow C_{min}=0\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

\(D=\left|x-2\right|+\left|x-8\right|\)

\(=\left|x-2\right|+\left|8-x\right|\ge6\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x-2\ge0\\x-8\ge0\end{cases}}\Rightarrow2\le x\le8\)