\(\left|x-1\right|+2C=\left|x-1,5\right|+\left|1-x\right|\\ \Leftrightarrow\left|x-1\right|+2C=\left|x-1,5\right|+\left|x-1\right|\\ \Rightarrow2C=\left|x-1,5\right|\ge0\\ \Rightarrow C\ge0\)

Để C=0 thì

\(\left|x-1,5\right|=0\\ \Leftrightarrow x-1,5=0\\ \Leftrightarrow x=1,5\)

Vậy...

Th1
x-1,5+2,5-x=0
<=>0x=-1
=> vô lí
Th2
-x+1,5-2,5+x=0
<=>0x=1
=>vô lí

vậy không có x thỏa mãn

x+1 mu 3 chu ko phai mu 2 nha tui ghi lon cai dau tien

Ta có: (x + 1)³ + (x - 2)³ - 2x² (x - 1,5) = 3

x³ + 1 + x³ - 8 - 2x³ + 3x² = 3

(x³ + x³ - 2x³ ) + 3x² +(1 - 8) = 3

3x² - 7 = 3

3x² = 10

x² = \(\frac{10}{3}\)

Vậy x = \(\sqrt{\frac{10}{3}}\)

4,75 x Y +5,43 xY =1,5

y  x ( 4,75 +5,43 ) =1,5

y x 10 ,18 = 1,5

tự chia tiếp nhé

\(\frac{-2}{1,5}=\frac{-2.2}{1,5.2}=\frac{-4}{3}=\frac{x-1}{3}\)

\(\Rightarrow x-1=-4\)\(\Rightarrow x=-3\)

Vậy \(x=-3\)

Ta có: \(-\left|1,5-x\right|\le0\forall x\)

\(\Rightarrow-\left|1,5-x\right|-2\le-2\forall x\)

Dấu \("="\) xảy ra khi \(\left|1,5-x\right|=0\)

\(\Rightarrow1,5-x=0\Rightarrow x=1,5\)

Vậy \(Min_A=-2\) khi \(x=1,5.\)

-2

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