|3x-4|=|x+2|

\(\Leftrightarrow\int^{3x-4=x+2}_{3x-4=-x-2}\Leftrightarrow\int^{3x-x=4+2}_{3x+x=4-2}\Leftrightarrow\int^{2x=6=>x=3}_{4x=2=>x=2}\)

vậy x E {2'3}

kết quả là 3
mình thi rồi, 300/300 đó

kết quả là 3
mình thi rồi, 300 đó

Ix+2I=Ix-2I
  =>    x+2=x-2    hoặc    x+2=2-x
          x-x =-2-2              x+x =2-2
           0x =4                    2x =0
 =>k có g/trị x TM               x =0
  Vậy x=0

|x+2|=|x-2|

=> x = 0

Ix+2I=Ix‐2I

=> x+2=x‐2 hoặc    x+2=2‐x

    x‐x =‐2‐ 2          x+x =2‐2

    0x =4                   2x =0

=>k có g/trị xTM        x =0

Vậy x=0 

a: \(A=\dfrac{x^2-8x+16-x^2+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-4x}{\left(x+4\right)\left(x-1\right)}\)

a: A=−4x/(x+4)(x−1)

Bài 1:

\(A=\left|x-3\right|+\left|x-5\right|+\left|x-7\right|\)

\(\ge x-3+0+7-x=4\)

Dấu = khi \(\begin{cases}x-3\ge0\\x-5=0\\7-x\le0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge3\\x=5\\x\le7\end{cases}\)\(\Leftrightarrow x=5\)

Vậy Min_A=4 khi x=5

Bài 2:

\(B=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-5\right|\)

\(\ge x-1+x-2+3-x+5-x=5\)

Dấu = khi \(\begin{cases}x-1\ge0\\x-2\ge0\\3-x\ge0\\5-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x\ge2\\x\le3\\x\le5\end{cases}\)\(\Leftrightarrow2\le x\le3\)

1) Ta có \(\hept{\begin{cases}\left|x\right|\ge0\forall x\\\left|y-2\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|x\right|+\left|y-2\right|\ge0\forall x;y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)

Vậy x  = 0 ; y = 2

Thay x = 0 ; y = 2 vào B 

=> B = 2.0 - 5.2 + 7.0.2 = -10

Vậy B = -10

Bài 2:

\(a)\)

\(A=\left|x-2021\right|+5\)

Ta có:

\(\left|x-2021\right|\ge0\Rightarrow\left|x-2021\right|+5\ge5\)

Dấu '' = '' xảy ra khi:

 \(x-2021=0\)

\(\Leftrightarrow x=2021\)

Vậy \(MinA=5\Leftrightarrow x=2021\)

\(b)\)

\(B=\left|x-2\right|+\left|x-5\right|\)

\(B=\left|x-2\right|+\left|x-5\right|\ge\left|x-2+5-x\right|=\left|3\right|=3\)

Dấu '' = '' xảy ra khi: 

\(\left(x-2\right)\left(5-x\right)\ge0\)

\(\Leftrightarrow2\le x\le5\)

Vậy \(MinB=3\Leftrightarrow2\le x\le5\)

2) a Ta có \(\left|x-2021\right|\ge0\forall x\)

=> \(\left|x-2021\right|+5\ge5\)

Dấu "=" xảy ra <=> x - 2021 = 0 

=> x = 2021

Vậy Min A = 5 <=> x = 2021

b) Ta có B = |x - 2| + |x - 5| = |x - 2| + |5 - x| \(\ge\left|x-2+5-x\right|=\left|3\right|=3\)

Dấu "=" xảy ra <=> (x - 2)(5 - x) \(\ge0\)

TH1 : \(\hept{\begin{cases}x-2\ge0\\5-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2\\x\le5\end{cases}}\Rightarrow2\le x\le5\)

TH2 : \(\hept{\begin{cases}x-2\le0\\5-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le2\\x\ge5\end{cases}}\Leftrightarrow x\in\varnothing\)

Vậy Min B = 3 <=> \(2\le x\le5\)