a |x + $\frac{1}{2}$ | + |x – $\frac{1}{3}$ | = 0
b |2x – $\frac{2}{7}$ | – |x + $\frac{2}{3}$ =0
a |x + $\frac{1}{2}$ | + |x – $\frac{1}{3}$ | = 0 b |2x – $\frac{2}{7}$ | – |x + $\frac{2}{3}$ =0
By Mary
By Mary
a |x + $\frac{1}{2}$ | + |x – $\frac{1}{3}$ | = 0
b |2x – $\frac{2}{7}$ | – |x + $\frac{2}{3}$ =0
Đáp án:
Giải thích các bước giải:
a ) Vì giá trị tuyệt đối của một số luôn luôn lớn hơn hoặc bằng 0
Mà | x + $\frac{1}{2}$ | + | x + $\frac{1}{3}$ | = 0
→ | x + $\frac{1}{2}$ | = | x + $\frac{1}{3}$ | = 0
Ta có : | x + $\frac{1}{2}$ | = | x + $\frac{1}{3}$ | ( Vô lí )
Vậy x ∈ ∅
b ) | 2x – $\frac{2}{7}$ | – | x + $\frac{2}{3}$ | = 0
→ | 2x – $\frac{2}{7}$ | = | x + $\frac{2}{3}$ |
TH 1 : 2x – $\frac{2}{7}$ = x + $\frac{2}{3}$
→ 2x = x + $\frac{20}{21}$
→ x = $\frac{20}{21}$
TH 2 : -( 2x – $\frac{2}{7}$ ) = -( x + $\frac{2}{3}$ )
→ -2x + $\frac{2}{7}$ = – x – $\frac{2}{3}$
→ -2x + $\frac{20}{21}$ = -x
→ -x + $\frac{20}{21}$ = 0
→ -x = $\frac{-20}{21}$
→ x = $\frac{20}{21}$
TH 3 : -( 2x – $\frac{2}{7}$ ) = x + $\frac{2}{3}$
→ -2x + $\frac{2}{7}$ = x + $\frac{2}{3}$
→ -2x = x + $\frac{8}{21}$
→ -3x = $\frac{8}{21}$
→ x = $\frac{-8}{63}$
TH 4 : 2x – $\frac{2}{7}$ = -( x + $\frac{2}{3}$ )
→ 2x – $\frac{2}{7}$ = -x – $\frac{2}{3}$
→ 2x + $\frac{8}{21}$ = -x
→ 3x + $\frac{8}{21}$ = 0
→ 3x = $\frac{-8}{21}$
→ x = $\frac{-8}{63}$
Vậy x = $\frac{-8}{63}$ và x = $\frac{20}{21}$