a) | 2,5 - x | = 1,3

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

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

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

còn lại tương tự

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

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

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

Vậy .

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

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

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

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

Vậy ....

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

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

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

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

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

Vậy ..........

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

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

Vậy .....

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

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

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

\(\Rightarrow x-0,2=0\)

\(\Rightarrow x=0,2\)

Vậy .......

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

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

Vậy .....

Chúc bạn học tốt!

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) \(1.6-\left|x-0.2\right|=0 \Leftrightarrow\left|x-0.2\right|=1.6\\ \Leftrightarrow\left\{{}\begin{matrix}x-0.2=1.6\\x-0.2=-1.6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1.8\\x=-1.4\end{matrix}\right.\)

c)\(\left|x-1.5\right|+\left|2.5-x\right|=0\)

Vì \(\left|x-1.5\right|\ge0\) và \(\left|2.5-x\right|\ge0\)

\(\Rightarrow\left|x-1.5\right|+\left|2.5-x\right|>0\) với mọi x

Vậy x ko có.

x:0.25-x:2.5=1.8
x:(0.25+0.5)=1.8
x:2.75=1.8
x=1.8x2.75
x=4.95
Lưu ý:      x:(0.25+0.5)=1.8    là 1 giả thuyết để bài làm đúng hơn còn trong bài tập các bạn làm mà không được giả thuyết thì đây được coi là 1 kết quả sai 

\(a,\Leftrightarrow25x^2-70x+49-25x^2=32\\ \Leftrightarrow-70x=-17\Leftrightarrow x=\dfrac{17}{70}\\ b,\Leftrightarrow x^2-6x+9+x^2+2x+1-5=0\\ \Leftrightarrow2x^2-4x+5=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+3=0\\ \Leftrightarrow2\left(x-1\right)^2=-3\Leftrightarrow\left(x-1\right)^2=-\dfrac{3}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)

\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)

\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)

\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)

hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)

Đơn thức là: x/(3y), x, -2/3, xy-2, 0

Đơn thức là \(-5xy^2;x;-\dfrac{2}{3};0\)

a. \(x^2-25-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b. \(\left(3x+1\right)^2=\left(2x-5\right)\\ \Leftrightarrow9x^2+6x+1=2x-5\\ \Leftrightarrow9x^2+6x-2x=-5-1\\ \Leftrightarrow9x^2+4x=-6\\ \Leftrightarrow x\left(9x+4\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\9x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{10}{9}\end{matrix}\right.\)

c. \(2x^2-7x+6=0\\ \Leftrightarrow2x^2-7x=-6\\ \Leftrightarrow x\left(2x-7\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{2}\end{matrix}\right.\)

a, \(\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=-2;x=5\)

b, bạn ktra lại đề, thường thường ngta hay cho 2 vế cùng bình phương 

c, \(2x^2-7x+6=0\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\Leftrightarrow x=\dfrac{3}{2};x=2\)

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

\(\left[{}\begin{matrix}3x+1=2x-5\\3x+1=5-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\5x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{4}{5}\end{matrix}\right.\)

\(\left(x\times1,5\right):2,75=1,25\)(dư 0,2)
\(x\times1,5=1,25\times2,75+0,2\)
\(x\times1,5=3,6375\)
\(x=3,6375:1,5\)
\(x=2,425\)

(\(x\) \(\times\) 1,5): 2,75 = 1,25 ( dư 0,2)

\(x\) \(\times\) 1,5  = 1,25 \(\times\) 2,75 + 0,2

\(x\) \(\times\)1,5  = 3,6375

\(x\) = 3,6375: 1,5

\(x\) = 2,425