a) x - 10 - (- 12) = 4

x-10=4+(-12)

x-10=-8

x=-8+10

x=2

=>giá trị tuyệt đối của x - 10 - (- 12) = 4 =/2/=2

b) 1

c) 2

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a) 2

b) 1

c) 2

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Bài 1:

\(a)\left(\dfrac{-28}{29}\right).\left(\dfrac{-38}{16}\right)=\dfrac{\left(-28\right).\left(-38\right)}{29.16}=\dfrac{1064}{464}=\dfrac{133}{58}\)

\(b)\left(\dfrac{-21}{16}\right).\left(\dfrac{-24}{7}\right)=\dfrac{\left(-21\right).\left(-24\right)}{16.7}=\dfrac{504}{112}=\dfrac{9}{2}\)

\(c)\left|\dfrac{-12}{17}\right|.\left(\dfrac{-34}{9}\right)=\dfrac{12}{17}.\left(\dfrac{-34}{9}\right)=\dfrac{12.\left(-34\right)}{17.9}=\dfrac{-408}{153}=\dfrac{-8}{3}\)

Bài 3:

\(a)\left|x\right|=21\)

\(\Rightarrow\left[{}\begin{matrix}x=-21\\x=21\end{matrix}\right.\)

\(b)\left|x\right|=\dfrac{17}{9};x< 0\)

\(\Rightarrow x=\dfrac{-17}{9}\)

\(c)\left|x\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(\left|x\right|=\dfrac{2}{5}\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)

\(d)\left|x\right|=0,35;x>0\)

\(\Rightarrow x=0,35\)

Bài 4:

\(a)\left|x\right|-1,7=2,3\)

\(\Rightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-3}{5}\end{matrix}\right.\)

\(b)\left|x\right|+\dfrac{3}{4}-\dfrac{1}{3}=0\)

\(\Rightarrow\left|x\right|+\dfrac{3}{4}=0+\dfrac{1}{3}\)

\(\Rightarrow\left|x\right|+\dfrac{3}{4}=\dfrac{1}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=\dfrac{-1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{-13}{12}\end{matrix}\right.\)

Chúc bạn học tốt!

\(\left|x^2-4x\right|+\left|x-2\right|\ge\left|x^2-4x+x-2\right|=\left|x^2-3x-2\right|\)

Dấu "=" xảy ra khi và chỉ khi \(\left(x^2-4x\right)\left(x-2\right)\ge0\)

\(\Leftrightarrow x\left(x-4\right)\left(x-2\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}0\le x\le2\\x\ge4\end{matrix}\right.\)

Vậy nghiệm của pt đã cho là \(\left[{}\begin{matrix}0\le x\le2\\x\ge4\end{matrix}\right.\)

Ta có : |5x + 1| + |3 - 2x|  \(\ge\left|5x+1+3-2x\right|=\left|4+3x\right|\)

Dấu "=" xảy ra <=> \(\left(5x+1\right)\left(3-2x\right)\ge0\)

Xét các trường hợp

TH1 : \(\hept{\begin{cases}5x+1\ge0\\3-2x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge-0,2\\x\le1,5\end{cases}}\Rightarrow-0,2\le x\le1,5\)

TH2 :\(\hept{\begin{cases}5x+1\le0\\3-2x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le-0,2\\x\ge1,5\end{cases}}\Rightarrow x\in\varnothing\)

Vậy  \(-0,2\le x\le1,5\)là giá trị cần tìm 

| x | - | 2 | = 5

=> | x | - 2 = 5

=> | x \ = 7

=> \(\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

3 | x | = 18

=> | x | = 6

=> \(\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

2 | x | - 5 = 7

=> | x | = 7 + 5 

=> | x | = 12

=> \(\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

| x | : 3 - 1 = | - 4 |

=> | x | : 3 - 1 = 4

=> | x | : 3 = 5

=> | x | = 15

=> \(\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)