Tìm x thuộc Qa) 3-4.\(\left|5-6x\right|\) =7b)\(\left|x+2\right|\) + \(\left|x+\frac{3}{5}\right|\) +\(\left|x+\frac{1}{2}\right|\) =4xc)\(\left|x+\frac{1}{101}\right|\) +\(\left|x+\frac{2}{101}\right|\) +\(\left|x+\frac{3}{101}\right|\) +...+\(\left|x+\frac{100}{101}\right|\) =101xlm giúp mk với, mk cần...
Đọc tiếp
Tìm x thuộc Q
a) 3-4.\(\left|5-6x\right|\) =7
b)\(\left|x+2\right|\) + \(\left|x+\frac{3}{5}\right|\) +\(\left|x+\frac{1}{2}\right|\) =4x
c)\(\left|x+\frac{1}{101}\right|\) +\(\left|x+\frac{2}{101}\right|\) +\(\left|x+\frac{3}{101}\right|\) +...+\(\left|x+\frac{100}{101}\right|\) =101x
lm giúp mk với, mk cần gấp
a ) \(3-4.\left|5-6x\right|=7\)
\(\Leftrightarrow4.\left|5-6x\right|=-4\)
\(\Leftrightarrow\left|5-6x\right|=-1\)
\(\Leftrightarrow\) Không thõa mãn ( vì \(x\ge0\) )
b) Do \(\left|x+2\right|\ge0;\left|x+\frac{3}{5}\right|\ge0;\left|x+\frac{1}{2}\right|\ge0\)
=> \(4x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)
=> \(\left(x+x+x\right)+\left(2+\frac{3}{5}+\frac{1}{2}\right)=4x\)
=> \(3x+\frac{31}{10}=4x\)
=> \(4x-3x=\frac{31}{10}\)
=> \(x=\frac{31}{10}\)
Vậy \(x=\frac{31}{10}\)
c) Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)
=> \(101x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 số x
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
Vậy x = 50