a)\(\left[{}\begin{matrix}\dfrac{5}{2}-x=\dfrac{1}{3}\\\dfrac{5}{2}-x=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{17}{6}\end{matrix}\right.\)

b) 8/6-x-1/5=0

9/6-x=1/5

x=13/10

a = 1,2

b = 1,8

a) \(\left|2,5-x\right|=1,3\)

\(\Leftrightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)

b) \(1,6-\left|x-0,2\right|=0\)

\(\Leftrightarrow\left|x-0,2\right|=1,6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

a) \(\left|2,5-x\right|-1,3=0\)

th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)

th2: \(2,5-x< 0\Leftrightarrow x>2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)

vậy \(x=1,2;x=3,8\)

b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)

c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)

th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)

th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)

vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)

d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)

th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)

vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)

e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm

a) \(\left|3,5-x\right|=1,3\)

\(\Rightarrow\left[{}\begin{matrix}3,5-x=1,3\\3,5-x=-1,3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3,5-1,3\\x=3,5+1,3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2,2\\x=4,8\end{matrix}\right.\)

b) \(1,6-\left|x-0,2\right|=0,4\)

\(\Rightarrow\left|x-0,2\right|=1,2\)

\(\Rightarrow\left[{}\begin{matrix}x-0,2=1,2\\x-0,2=-1,2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1,2+0,2\\x=-1,2+0,2\end{matrix}\right.\)

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

\(\left|3,5-x\right|=1,3\)

\(\Rightarrow\left[{}\begin{matrix}3,5-x=1,3\Rightarrow x=2,2\\3,5-x=-1,3\Rightarrow x=4,8\end{matrix}\right.\)

\(1,6-\left|x-0,2\right|=0,4\)

\(\Rightarrow\left|x-0,2\right|=1,2\)

\(\Rightarrow\left[{}\begin{matrix}x-0,2=1,2\Rightarrow x=1,4\\x-0,2=-1,2\Rightarrow x=-1\end{matrix}\right.\)

\(\left|x-1,5\right|+\left|2,5-x\right|=0\)

\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-1,5\right|=0\Rightarrow x=1,5\\\left|2,5-x\right|=0\Rightarrow x=2,5\end{matrix}\right.\)

\(1,5\ne2,5\Rightarrow x\in\varnothing\)

c) \(\left|x-1,5\right|+\left|2,5-x\right|=0\)

Ta có : \(\left|x-1,5\right|\ge0\) với mọi \(x\)

\(\left|2,5-x\right|\ge0\) với mọi \(x\)

Nên \(\left|x-1,5\right|+\left|2,5-x\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

Vậy không tìm được giá trị thõa mãn của \(x\)

Chúc học tốt !!!

Bài 1:a/ 1.6-Ix-0.2I=0

Có 2 trường hợp:

TH1: x-0.2=1.6

=> x=1.6+0.2=1.8

TH2: x-0.2=-1.6

=> x=-1.4

b/ Có 2 trường hợp:

TH1:x-1.5=0=>x=1.5

TH2: 2.5-x=0=> x=2.5

Bài 2: a/ Vì Ix-3.5I\(\ge0\)

=> A_max=0.5-0=0.5 khi x=3.5

          b/ Vì -I1.4-xI \(\le0\)

Nên B_max=0-2=-2 khi x=1.4

|2,5-x|=1,3

\(\orbr{\begin{cases}2,5-x=1,3\\2,5-x=-1,3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)

Vậy x=1,2 hoặc x=3,8

|x-1,5|+|2,5-x|=0

\(\Rightarrow\hept{\begin{cases}VT:x-1,5=0\\VP:2,5-x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1,5\\x=2,5\end{cases}}\)

Vậy x của VT là 1,5 và x của VP là 2,5

\(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\)

x=\(0+\frac{1}{2}\)

x=\(\frac{1}{2}\)

(x-2)²=1

=> x-2=1

x=1+2

x=3

=> x-2=-1

x=(-1)+2

x=1

a, / 2,5 - x / = 1,3

Với 2,5 - x > hoặc = 0 => 2, 5 - x  = 1,3

                         => x = 1, 2

Với 2,5 - x < hoặc = 0 =>  - ( 2,5 - x ) = 1,3

                                    => - 2,5 + x = 1,3

                                    => x =  3,8

Vậy x  thuộc tập hợp 1,2 ; 3,8

     p/s: > hoặc = 0,  < hoặc = 0 ,  thuộc tập hợp bạn ghi kí hiệu nha

a) \(\left(x-1,3\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-1,3=3\\x-1,3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\)

b) 2^4-x = 32

⇔ 2^4-x = 2⁵

⇔ 4-x=5

⇔ x=-1

c) (x+1,5)²+(y-2,5)¹⁰=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1,5\\y=2,5\end{matrix}\right.\)

\(a,\left(x-1,3\right)^2=9\\ \Leftrightarrow\left(x-1,3+9\right)\left(x-1,3-9\right)=0\\ \Leftrightarrow\left(x-7,7\right)\left(x-10,3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7,7=\dfrac{77}{10}\\x=10,3=\dfrac{103}{10}\end{matrix}\right.\)

\(b,2^{4-x}=32=2^5\\ \Leftrightarrow4-x=5\\ \Leftrightarrow x=-1\)

\(c,\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+1,5=0\\y-2,5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1,5=-\dfrac{3}{2}\\y=2,5=\dfrac{5}{2}\end{matrix}\right.\)

a. (x - 1,3)² = 9

<=> (x - 1,3)² - 9 = 0

<=> (x - 1,3)² - 3² = 0

<=> (x - 1,3 - 3)(x - 1,3 + 3) = 0

<=> (x - 4,3)(x + 1,7) = 0

<=> \(\left[{}\begin{matrix}x-4,3=0\\x+1,7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4,3\\x=-1,7\end{matrix}\right.\) 

\(\left[{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\left[{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)

\(\left[{}\begin{matrix}x-\dfrac{4}{5}=\dfrac{3}{4}\\x-\dfrac{4}{5}=\dfrac{-3}{4}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{20}\\x=\dfrac{1}{20}\end{matrix}\right.\)

c.x thuộc tập hợp rỗng

d. cậu chia ra thành 2 trường hợp nhé