Toán Lớp 7: a, |x-1| – $\frac{2}{5}$ = $\frac{-1}{4}$
b, (2x+1)³=64
c, ($\frac{2}{5}$)^x – $\frac{7}{5}$= $\frac{-27}{125}$
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Toán Lớp 7: a, |x-1| – $\frac{2}{5}$ = $\frac{-1}{4}$
b, (2x+1)³=64
c, ($\frac{2}{5}$)^x – $\frac{7}{5}$= $\frac{-27}{125}$
Comments ( 2 )
a) |x – 1| – $\frac{2}{5}$ = $\frac{-1}{4}$
|x – 1| = $\frac{-1}{4}$ + $\frac{2}{5}$
|x – 1| = $\frac{-5}{20}$ + $\frac{8}{20}$
|x – 1| = $\frac{3}{20}$
TH1: x – 1 = $\frac{3}{20}$
x = $\frac{3}{20}$ + 1
x = $\frac{3}{20}$ + $\frac{20}{20}$
x = $\frac{23}{20}$
TH2: x – 1 = $\frac{-3}{20}$
x = $\frac{-3}{20}$ + 1
x = $\frac{-3}{20}$ + $\frac{20}{20}$
x = $\frac{17}{20}$
Vậy x ∈ { $\frac{23}{20}$ ; $\frac{17}{20}$ }
b) (2x + 1)³ = 64
(2x + 1)³ = 4³
2x + 1 = 4
2x = 4 – 1
2x = 3
x = 3 : 2
x = 1,5
Vậy x = 1,5
c) $(\frac{2}{5})^{x}$ – $\frac{7}{5}$ = $\frac{-27}{125}$
$(\frac{2}{5})^{x}$ = $\frac{-27}{125}$ + $\frac{7}{5}$
$(\frac{2}{5})^{x}$ = $\frac{-27}{125}$ + $\frac{175}{125}$
$(\frac{2}{5})^{x}$ = $\frac{148}{125}$
x ∈ ∅
a, |x – 1| – 2/5 = (-1)/4
|x – 1| = (-1)/4 + 2/5
|x – 1| = (-5)/20 + 8/20
|x – 1| = 3/20
Ta có:
TH1:
x – 1 = 3/20
x = 3/20 + 1
x = 23/20
TH2:
x – 1 = (-3)/20
x = (-3)/20 + 1
x = 17/20
Vậy x = 23/20 hoặc x = 17/20
b, (2x + 1)^3 = 64
(2x + 1)^3 = 4^3
2x + 1 = 4
2x = 3
x = 3/2
Vậy x = 3/2
c, (2/5)^x – 7/5 = (-27)/125
(2/5)^x = (-27)/125 + 7/5
(2/5)^x = (-27)/125 + 175/125
(2/5)^x = 148/125
x in ∅
Vậy x không có giá trị thỏa mãn