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\(-\frac{5}{2}< \frac{3}{x}< -\frac{2}{3}\)
\(\Rightarrow-\frac{15}{6}< \frac{3}{x}< -\frac{4}{6}\)
\(\frac{3}{x}\in\left(-\frac{5}{6};-1;-\frac{7}{6};-\frac{4}{3};-\frac{3}{2};-\frac{5}{3};-\frac{11}{6};-2;-\frac{13}{6};-\frac{7}{3};\right)\)
\(\Rightarrow x\in\left(-2;-3\right)\)
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\(A=\left(a+b\right)-\left(-a-b\right)+\left(a-c\right)-\left(a+c\right)\)
\(A=a+b+a+b+a-c-a-c\)
\(A=\left(a+a+a-a\right)+\left(b+b\right)+\left(-c-c\right)\)
\(A=3a+2b-2c\)
\(B=\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)\)
\(B=a+b-c+a-b+c-b-c+a-a+b+c\)
\(B=\left(a+a+a-a\right)+\left(b-b-b+b\right)+\left(-c+c-c+c\right)\)
\(B=3a\)
a) \(A=\left(a+b\right)-\left(-a-b\right)+\left(a-c\right)+\left(a+c\right)\)
\(=a+b+a+b+a-c-a-c=2a+2b-2c=2\left(a+b-c\right)\)
b) \(B=\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)\)
\(=a+b-c+a-b+c-b-c+a-a+b+c\)
\(=3a\)
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\(9x-5x+3x=56\)
<=> \(7x=56\)
<=> \(x=8\)
Vậy...
\(9x-x+3x=165\)
<=> \(11x=165\)
<=> \(x=15\)
Vậy....
\(7x+8x-x=168\)
<=> \(14x=168\)
<=> \(x=12\)
Vậy...
\(\frac{5}{8}x-\frac{1}{3}x-\frac{1}{6}x=15\)
<=> \(\frac{1}{8}x=15\)
<=> \(x=120\)
Vậy...
Dấu . là dấu nhân nha
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a, 9 . x - 5 . x + 3 . x = 56
x . (9 - 5 + 3) = 56
x . 7 = 56
x = 56 : 7
x = 8
b, 9 . x - x + 3 . x = 165
x . (9 - 1 + 3) = 165
x . 11 = 165
x = 165 : 11
x = 15
c, 7 . x + 8 . x - x = 168
x . (7 + 8 - 1) = 168
x . 14 = 168
x = 168 : 14
x = 12
d, 5/8 . x - 1/3 . x - 1/6 . x = 15
x . (5/8 - 1/3 - 1/6) = 15
x . 1/8 = 15
x = 15 : 1/8
x = 120
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vay mik ket ban voi bn thi ban cung phai tra loi nhung cau hoi mik dang len
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mình hâm mộ anh Nguyễn Thành Nam,đặc biệt là kênh monter ntn
kb với mk nha
k tớ...Nha!