Question

Tìm x: \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

Answer 1

148-x/25-1 + 169-x/23-2 + 186-x/21-3 + 199-x/19-4
123-x/25 + 123-x/23 + 123-x/21 + 123-x/19 =0

123-x=0 => x=123

Answer 2

\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)

\(\left(\frac{148-x}{25}-1\right)+\left(\frac{169-x}{23}-2\right)+\left(\frac{186-x}{21}-3\right)+\left(\frac{199-x}{19}-4\right)=0\)

=> \(\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)

=> \(\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

=> 123 - x = 0

=> x = 123

Answer 3

duongtiendung vế bên trái có thêm -1,-2,-3,-3 thì bên vế phải ,phải là 0+(-1)+(-2)+(-3)+(-4) 

=-10 chứ = 0 sao đc