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a) \(\left[0,\left(37\right)+0,\left(62\right)\right]\cdot x=10\)
=> \(\left[\frac{37}{99}+\frac{62}{99}\right]\cdot x=10\)
=> \(1\cdot x=10\Rightarrow x=10\)
b) \(\frac{0,\left(12\right)}{1,\left(6\right)}=\frac{\frac{12}{99}}{\frac{5}{3}}=\frac{12}{99}\cdot\frac{3}{5}=\frac{4}{55}\)
=> \(\frac{4}{55}=x:0,\left(4\right)\)
=> \(\frac{4}{55}=x:\frac{4}{9}\)
=> \(x:\frac{4}{9}=\frac{4}{55}\)
=> \(x=\frac{4}{55}\cdot\frac{4}{9}=\frac{16}{495}\)
B1 :
\(\frac{0,1\left(6\right)+0,\left(3\right)}{0,\left(3\right)+1,1\left(6\right)}\) . x = 0,(2)
=\(\frac{0,5}{1,5}\).x=0,(2)
x=0,(2):\(\frac{0,5}{1,5}\)
x=0,(6)=\(\frac{2}{3}\)
b2:
[12,(1) - 2,3(6)] : 4,(21)
=9,7(4):4,(21)
=\(\frac{9,7\left(4\right)}{4,\left(21\right)}\)
\(\dfrac{12}{99}\):(x-1)=\(\dfrac{5}{3}\):\(\dfrac{4}{9}\)
x-1=\(\dfrac{16}{495}\)
x=\(\dfrac{511}{495}\)
\(0,\left(12\right):\left(x-1\right)=1,\left(6\right):0,\left(4\right)\)
\(\Rightarrow\dfrac{12}{99}:\left(x-1\right)=\dfrac{5}{3}:\dfrac{4}{9}\)
\(\Rightarrow\dfrac{4}{33}:\left(x-1\right)=\dfrac{5}{3}.\dfrac{9}{4}\)
\(\Rightarrow\dfrac{4}{33}:\left(x-1\right)=\dfrac{15}{4}\)
\(\Rightarrow x-1=\dfrac{4}{33}:\dfrac{15}{4}\)
\(\Rightarrow x-1=\dfrac{16}{495}\)
\(\Rightarrow x=\dfrac{16}{495}+1\Rightarrow x=\dfrac{511}{495}\)
Còn cách biến đổi làm sao thì bạn xem ở đây nhá https://diendan.hocmai.vn/threads/doi-so-thap-phan-1-4-51-ra-phan-so-toi-gian.393734/
Cho ví dụ luôn biến đổi \(1,\left(6\right)\) sang phân số là: \(1,\left(6\right)=1+0,\left(6\right)=1+\dfrac{2}{3}=\dfrac{3}{3}+\dfrac{2}{3}=\dfrac{5}{3}\)
Chúc bạn học tốt 
\(\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|+20x=0\left(1\right)\)
TH1: x<1
(1) trở thành 1-x+2(2-x)+3(3-x)+4(4-x)+5(5-x)+20x=0
=>\(1-x+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(5x+55=0\)
=>x=-11(nhận)
TH2: 1<=x<2
Phương trình (1) sẽ trở thành:
\(x-1+2\left(2-x\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(7x+53=0\)
=>\(x=-\dfrac{53}{7}\left(loại\right)\)
TH3: 2<=x<3
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+9-3x+16-4x+25-5x+20x=0\)
=>\(11x+45=0\)
=>\(x=-\dfrac{45}{11}\left(loại\right)\)
TH4: 3<=x<4
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(x-3\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+16-4x+25-5x+20x=0\)
=>\(-3x+27=0\)
=>x=9(loại)
TH5: 4<=x<5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+25-5x+20x=0\)
=>\(25x-5=0\)
=>x=1/5(loại)
TH6: x>=5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(x-5\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+5x-25+20x=0\)
=>35x-55=0
=>x=55/35(loại)
a, \(\Leftrightarrow x^2+2x+1+\left|x+10\right|-x^2-12=0\)
\(\Leftrightarrow\left|x+10\right|+2x-11=0\)
ta có ; | x+10| = x+10 khi x+10\(\ge\)0 hay x \(\ge\)-10
|x+10| = -x-10 khi x+10<0 hay x<-10
vs x\(\ge\)-10 ta có: x+10+2x-11=0 \(\Leftrightarrow\)3x=1 \(\Leftrightarrow\)x= \(\frac{1}{3}\)( thỏa mãn )
vs x< -10 ta có (tự thay vào r tính típ)
vậy x=...............
b, lm tg tự
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
0,(12) : (x - 1) = 1,(6) : 0,(4)
=> 0,(1) .12 : (x - 1) = [0,(1) . 6 + 1] : [0,(1) . 4]
=> 1/9 . 12 :(x - 1) = (1/9 . 6 + 1) : (1/9 . 4)
=> 4/3 :(x - 1) = 5/3 : 4/9
=> 4/3 : (x - 1) = 15/4
=> x - 1 = 4/3 : 15/4
=> x - 1 = 16/45
=> x = 16/45 + 1
=> x = 61/45