Bài làm

2009 - | - 2009 | = x

2009 -     2009   = x

           0             = x

Vậy x = 0

# Học tốt #

2009+|x-2009|=0

|x-2009|=-2009

mà |x-2009| \(\ge\)0

\(\Rightarrow\)không tìm được giá trị nào của x thỏa mãn

/x-2009/=2009-x\(\Rightarrow\)x-2009< hoạc = 0\(\Rightarrow\)x< hoạc = 2009

\(2009-\left|x-2009\right|=x\Rightarrow\left|x-2009\right|=2009-x\)

ĐK: \(x\ge2009\)

\(\Rightarrow\orbr{\begin{cases}x-2009=2009-x\\x-2009=x-2009\end{cases}\Rightarrow\orbr{\begin{cases}x=2009\left(tm\right)\\0=0\left(loại\right)\end{cases}}}\)

ta có:

\(\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|x-4\right|\)

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

\(\ge\left|x-1+4-x\right|+\left|x-2\right|+\left|y-3\right|\)

\(=3+\left|x-2\right|+\left|y-3\right|\)

\(\ge3\)

Dấu "=" xả ra khi \(\hept{\begin{cases}\left(x-1\right)\left(4-x\right)\ge0\\\left|x-2\right|=0\\\left|y-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}1\le x\le4\cdot\\x=2\left(TM\cdot\right)\\y=3\end{cases}}\)

Vậy \(x=2;y=3\)

(x-1) + (x-2) + (x-3) + (x-4) = 3

(x+x+x+x) - (1+2+3+4) = 3

X x 4 - 10 = 3

X x 4 = 3 + 10

X x 4 = 13

x = 13 : 4

x = \(\frac{13}{4}\)

Ta có:\(\left|x+1\right|\ge0;\left|x-2\right|\ge0;\left|x+7\right|\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x-2\right|+\left|x+7\right|\ge0\)

\(\Rightarrow5x-10\ge0\)

\(\Rightarrow5x\ge10\)

\(\Rightarrow x\ge2\)

\(\Rightarrow\left|x+1\right|=x+1\)

  \(\left|x-2\right|=x-2\)

  \(\left|x+7\right|=x+7\)

Ta có:\(\left|x+1\right|+\left|x-2\right|+\left|x+7\right|=5x-10\)

\(\Rightarrow x+1+x-2+x+7=5x-10\)

\(\Rightarrow\)\(3x+6=5x-10\)

\(\Rightarrow6+10=5x-3x\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=8\)

Vậy x=8 thỏa mãn

\(Ix^2+Ix-1II=x^2+2\Leftrightarrow x^2+Ix-1I=x^2+2\Rightarrow Ix-1I=2\)

\(\orbr{\begin{cases}x-1=2=>x=3\\x-1=-2=>x=-1\end{cases}}\)

Ta có : 

\(2009-\left|x-2009\right|=x\)

\(\Leftrightarrow\)\(\left|x-2009\right|=2009-x\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2009=x-2009\\x-2009=2009-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=x\\x=2009\end{cases}}}\)

Vậy \(x=2009\)

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