=>\(\left[\begin{array}{nghiempt}x-1,4=x+3,1\\x-1,4=-x-3,1\end{array}\right.\)

=>\(\left[\begin{array}{nghiempt}0x=4,5\\2x=-1,7\end{array}\right.\)

=>x=-0,85

|x-1,4|=|x+3,1|

\(\Rightarrow\orbr{\begin{cases}x-1,4=x+3,1\\x-1,4=-x-3,1\end{cases}}\Rightarrow\orbr{\begin{cases}x-x=3,1+1,4\\x+x=-3,1+1,4\end{cases}}\Rightarrow\orbr{\begin{cases}0x=4,5\\2x=-1,7\end{cases}}\Rightarrow\orbr{\begin{cases}loai\\x=-0,75\end{cases}}\)

Vậy x=-0,75

nếu x-1,4=x+3,1

0x=4,5

x không có giá trị

nếu x-1,4=-(x+3,1)

x-1,4=-x-3,1

2x= -1,7

x=-0,75(lấy)

vậy x=-0,75

\(\Leftrightarrow\left[{}\begin{matrix}x-1.4=x+3.1\\1.4-x=x+3.1\end{matrix}\right.\Leftrightarrow-2x=1.7\)

hay x=-1,7/2=-0,85

Theo bài ra ta có: \(\left|x-4\right|=\left|x+3,1\right|\)

=> x-4 = \(\pm\) (x+3,1)

=>\(\left[\begin{array}{nghiempt}x-4=x+3,1\\x-4=-x-3,1\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x-x=3,1+4\\x+x=-3,1+4\end{array}\right.\)

=> \(\left[\begin{array}{nghiempt}0=7,1\left(loại\right)\\2x=0,9\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x\in\varnothing\\x=0,45\end{array}\right.\)

Vậy x=0,45

Suy ra|x-1,4|=x+3,1 hoặc |x-1,4|=-(x+3,1)

Th1:|x-1,4|=x+3,1 thì x-1,4=x+3,1 hoặc -(x-1,4)=x+3,1

+) nếu x-1,4=x+3,1 thì:x-x=3,1+1,4=4,5 hay 0=4,5( vô lí)

+) nếu -(x-1,4)=x+3,1 thì-x-x=3,1-1,4 hay -2x=1,7 suy ra x=1,7:(-2)=-17/20

Th2:tương tự

-1,85 nha

Áp dụng \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) vào biểu thức

Xét VT ta có:\(\left|\left|x-12,2\right|+\left|x-22,6\right|\right|=\left|\left|x-12,2\right|+\left|22,6-x\right|\right|\)

Mà \(\left|x-12,2\right|+\left|22,6-x\right|\ge\left|x-12,2+22,6-x\right|=10,4\)

\(\Rightarrow\left|\left|x-12,2\right|+\left|x-22,6\right|\right|\ge10,4\)

Vì 10,4>10\(\Rightarrow\left|\left|x-12,2\right|+\left|x-22,6\right|\right|>10\)(vô lí vì VP=10)

Vậy x không có giá trị

0 gia tri

thanks bạn

\(x\left(x-2022\right)-2x+4044=0\\ \Rightarrow x\left(x-2022\right)-2\left(x-2022\right)=0\\ \Rightarrow\left(x-2022\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2022\\x=2\end{matrix}\right.\)

x(x-2022)-2x+4044=0

\(x^2-2022x-2x+4044=0\)

\(x^2-2x-2022x+4044=0\)

\(\left(x^2-2x\right)-\left(2022x-4044\right)=0\)

\(x\left(x-2\right)-2022\left(x-2\right)=0\)

\(\left(x-2\right)\left(x-2022\right)=0\)

\(x-2=0\)

\(x-2022=0\)

<=> \(x=\left\{2;2022\right\}\)