|\(x\)| - 2,2 = 1,3
|\(x\)| = 1,3 + 2,2
|\(x\)| = 3,5
\(\left[{}\begin{matrix}x=-3,5\\x=3,5\end{matrix}\right.\)
Vậy \(x\in\) {-3,5; 3,5}
\(\left|x\right|-2,2=1,3\)
\(\left|x\right|=1,3+2,2\)
\(\left|x\right|=3,5\)
\(x=-3,5\); \(x=3,5\)
|\(x\)| - 2,2 = 1,3
|\(x\)| = 1,3 + 2,2
|\(x\)| = 3,5
\(\left[{}\begin{matrix}x=-3,5\\x=3,5\end{matrix}\right.\)
Vậy \(x\in\) {-3,5; 3,5}
\(\left|x\right|-2,2=1,3\)
\(\left|x\right|=1,3+2,2\)
\(\left|x\right|=3,5\)
\(x=-3,5\); \(x=3,5\)
Tìm \(x\in Q\), biết:
a, \(\left|2,5-x\right|=1,3\)
b, \(1,6-\left|x-0,2\right|=0\)
Bài 1:
a) \(\left(x-1,3\right)^2=9\)
b) \(2^{4-x}=32\)
c) \(\left(x+1,5\right)^2+\left(y-2,5\right)^{10}=0\)
\(\left|x-2,2\right|=\left|0,2+x\right|\) Tìm x
\(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)
Tìm x biết:
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x-1\)-1
\(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(\left|x+1\right|-\left|3x+2\right|=x+2\)
/x/-2,2=1,3
Bài 2 : Rút gọn biểu thức sau :
a ) \(A=\left|x+1,3\right|-\left|x-2,5\right|.\)khi \(x< -1,3\)
b ) \(D=\left|x+3\frac{1}{2}\right|+\left|x\right|-3\frac{1}{2}\)với \(x>0\)
c ) \(A=\left|x-\frac{1}{7}\right|-\left|x+\frac{3}{5}\right|+\frac{4}{5}\)khi \(-\frac{3}{5}< x< \frac{1}{7}\)
tìm x,y thoả mãn:\(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)
\(\left|x-1,3\right|=2,7\)
\(\left|x+1,2\right|-2,6=4,5\)
a) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\\ \)
b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)
c) \(\left(0,4x-1,3\right)^2=5,29\)