tim x biet 0,2- |4,2-2x| =0
Tìm x
0,2 - /4,2 - 2x / = 0
\(0,2-\left|4,2-2x\right|=0\)
\(\Leftrightarrow\left|4,2-2x\right|=0,2\)
\(\Leftrightarrow\orbr{\begin{cases}4,2-2x=0,2\\4,2-2x=-0,2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=2,2\end{cases}}\)
Vậy \(x\in\left\{2;2,2\right\}\)
\(0,2-\left|4,2-2x\right|=0\)
\(\Rightarrow\left|\frac{21}{5}-2x\right|=\frac{1}{5}\)
\(\Rightarrow\orbr{\begin{cases}\frac{21}{5}-2x=\frac{1}{5}\\\frac{21}{5}-2x=-\frac{1}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=\frac{22}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{11}{5}\end{cases}}\)
\(0,2-\left|4,2-2x\right|=0\)
\(\Leftrightarrow\left|4,2-2x\right|=0,2\)
\(\Leftrightarrow\orbr{\begin{cases}4,2-2x=0,2\\4,2-2x=-0,2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4,2-0,2\\2x=4,2+0,2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=4,4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=2,2\end{cases}}\)
tim x biet
a) |0,2.x-3,1|=63
b) |0,2.x-3,1|+|0,2.x+3,1|=0
Tìm x
0,2 - | 4,2 - 2x | = 0
Giải dùm ạ
0,2 - | 4,2 - 2x | = 0
| 4,2 - 2x | = 0,2 - 0
| 4,2 - 2x| = 0,2
=> 4,2 - 2x = 0,2 hoặc 4,2 - 2x = -0,2
+> 4,2 - 2x = 0,2
2x = 4,2 - 0,2
2x = 4
x = 4:2
x = 2
+> 4,2 -2x = -0,2
2x = 4,2+ 0,2
2x = 4,4
x= 4,4 :2
x = 2,2
0,2 - | 4,2 - 2x | = 0
| 4,2 - 2x | = 0,2 - 0
| 4,2 - 2x | = 0,2
=> 4,2 - 2x = \(\pm\)0,2
+) 4,2 - 2x = 0,2 +) 4,2 - 2x = - 0,2
2x = 4,2 - 0,2 2x = 4,2 - (- 0,2 )
2x = 4 2x = 4,4
x = 4:2 =2 x = 4,4:2 = 2,2
Vậy x \(\in\){ 2; 2,2}
0,2 - /4,2-2x/=0
tim xy biet (x-0,2)^10+(y+3,1)=0
Tim x biet: 0,1 + 0,2 +...+ 0,x = 4,5
tim x biet x *5,2- x=4,2 *10
\(x\) \(\times\) 5,2 - \(x\) = 4,2 \(\times\) 10
\(x\) \(\times\) 5,2 - \(x\) \(\times\) 1 = 42
\(x\) \(\times\) ( 5,2 - 1) = 42
\(x\) \(\times\) 4,2 = 42
\(x\) = 42 : 4,2
\(x\) = 10
tinh
0,2-/4,2-2x/=0
0,2-|4,2-2x|=0
|4,2-2x|=0,2
=>4,2-2x=0,2
2x=4
x=2
=>4,2-2x=-0,2
2x=4,4
x=2,2
tìm x
a) |2x-1| =5
b) \(\left|x+\frac{3}{4}\right|=3\)
c) 0,2 - |4,2-2x| = 0
a) \(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\)
Vậy: \(x=3\) hoặc \(x=-3\)
b) \(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{3}{4}=3\\x+\frac{3}{4}=-3\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{9}{4}\\x=-\frac{15}{4}\end{array}\right.\)
Vậy: \(x=\frac{9}{4}\) hoặc \(x=-\frac{15}{4}\)
c) \(\Rightarrow\left[\begin{array}{nghiempt}4,2-2x=0,2\\4,2-2x=-0,2\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{11}{5}\end{array}\right.\)
Vậy: \(x=2\) hoặc \(x=\frac{11}{5}\)