Tìm x ∈ R biết
Huhuhuj mn ơi e mất gốc toán í ạa ;((
Ai giúp mình với, bnhieu câu cũng đc
Ai làm hết mình cưới nha <3
----- Nội dung ảnh -----
Bài 2: Tìm x ∈ R biết:
a) \( x + \frac{4}{3} - \frac{5}{4} = -\frac{3}{6} \)\
c) \( |2x + 1| + \frac{2}{3} - |x - \frac{1}{2}| = \frac{4}{3} \)\
e) \( |x + 2| - \frac{8}{3} = |\sqrt{4}| \)\
g) \( \frac{1}{2^{2 - |x|}} - \frac{3}{4} = \frac{2^{2 - 3}}{16} - 2 \)\
b) \( 5 \left( x + \frac{3}{5} \right) - 3 \left( x - \frac{1}{3} \right) = -x + \frac{1}{2} \)\
d) \( 5 - |x - \frac{1}{2}| = \frac{2}{3} \)\
f) \( |2x - 1| + |2| = \sqrt{4} \)\
h) \( \sqrt{\frac{1}{25}} + |x - 2| = 1.1 \)\
k) \( (2x + 1) \left( \frac{2}{3} x - 1 \right) = 0 \)\
m) \( (2x - 3)^2 = \frac{49}{81} \)\
n) \( \left( \frac{2}{3} + x \right)^2 = 0.25 \)\
l) \( \frac{1}{2} \left( x + \frac{1}{4} \right)^2 = \frac{17}{36} \)\
o) \( (2x + 1)^2 = \frac{8}{125} \)
Ai giúp mình với, bnhieu câu cũng đc
Ai làm hết mình cưới nha <3
----- Nội dung ảnh -----
Bài 2: Tìm x ∈ R biết:
a) \( x + \frac{4}{3} - \frac{5}{4} = -\frac{3}{6} \)\
c) \( |2x + 1| + \frac{2}{3} - |x - \frac{1}{2}| = \frac{4}{3} \)\
e) \( |x + 2| - \frac{8}{3} = |\sqrt{4}| \)\
g) \( \frac{1}{2^{2 - |x|}} - \frac{3}{4} = \frac{2^{2 - 3}}{16} - 2 \)\
b) \( 5 \left( x + \frac{3}{5} \right) - 3 \left( x - \frac{1}{3} \right) = -x + \frac{1}{2} \)\
d) \( 5 - |x - \frac{1}{2}| = \frac{2}{3} \)\
f) \( |2x - 1| + |2| = \sqrt{4} \)\
h) \( \sqrt{\frac{1}{25}} + |x - 2| = 1.1 \)\
k) \( (2x + 1) \left( \frac{2}{3} x - 1 \right) = 0 \)\
m) \( (2x - 3)^2 = \frac{49}{81} \)\
n) \( \left( \frac{2}{3} + x \right)^2 = 0.25 \)\
l) \( \frac{1}{2} \left( x + \frac{1}{4} \right)^2 = \frac{17}{36} \)\
o) \( (2x + 1)^2 = \frac{8}{125} \)