Toán tìm x a,60%x-x+1/3=2 b,x+1/7=1/2 c,(4x^2 – 25)(x+3)=0 18/10/2021 By Reese tìm x a,60%x-x+1/3=2 b,x+1/7=1/2 c,(4x^2 – 25)(x+3)=0
a) 60%x – x + $\frac{1}{3}$ = 2 $\frac{3}{5}$x – x + $\frac{1}{3}$ = 2 $\frac{3}{5}$x – 1x + $\frac{1}{3}$ = 2 x($\frac{3}{5}$ – 1) + $\frac{1}{3}$ = 2 x $\frac{-2}{5}$ + $\frac{1}{3}$ = 2 x $\frac{-2}{5}$ = 2 – $\frac{1}{3}$ x $\frac{-2}{5}$ = $\frac{4}{3}$ x = $\frac{4}{3}$ : $\frac{-2}{5}$ x = $\frac{-20}{6}$ b) x + $\frac{1}{7}$ = $\frac{1}{2}$ x = $\frac{1}{2}$ – $\frac{1}{7}$ x = $\frac{5}{14}$ Trả lời
b, x= 5/14
c, x= -3 hoặc x= 5/2 hoặc x= -5/2
Giải thích các bước giải:
a) 60%x – x + $\frac{1}{3}$ = 2
$\frac{3}{5}$x – x + $\frac{1}{3}$ = 2
$\frac{3}{5}$x – 1x + $\frac{1}{3}$ = 2
x($\frac{3}{5}$ – 1) + $\frac{1}{3}$ = 2
x $\frac{-2}{5}$ + $\frac{1}{3}$ = 2
x $\frac{-2}{5}$ = 2 – $\frac{1}{3}$
x $\frac{-2}{5}$ = $\frac{4}{3}$
x = $\frac{4}{3}$ : $\frac{-2}{5}$
x = $\frac{-20}{6}$
b) x + $\frac{1}{7}$ = $\frac{1}{2}$
x = $\frac{1}{2}$ – $\frac{1}{7}$
x = $\frac{5}{14}$