Tìm x
a) x ∈ BC (7;14;20) và 100 ≤ x ≤ 300, x ∈ N
b) x ⋮ 4; x ⋮ 7; x ⋮ 8 và x là số nhỏ nhất, x ∈ N
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a: x=9/2-3/7=57/14
b: =>x=7/5x5/7=1
c: =>x=11/3:3/8=11/3x8/3=88/9
a : x = \(\dfrac{9}{2}\) - \(\dfrac{3}{7}\) = \(\dfrac{57}{14}\)
b : = > x = \(\dfrac{7}{5}\) x\(\dfrac{5}{7}\) = 1
c : = > \(\dfrac{11}{3}\) : \(\dfrac{3}{8}\) = \(\dfrac{11}{3}\) x \(\dfrac{8}{3}\) = \(\dfrac{88}{9}\)
a.\(\left(-12\right)x-14=-2\)
\(\left(-12\right)x=-2+14\)
\(\left(-12\right)x=12\)
\(x=12:\left(-12\right)\)
\(x=-1\)
\(b,\left(-8\right)x=\left(-5\right)\left(-7\right)-3\)
\(\left(-8\right)x=35-3\)
\(\left(-8\right)x=32\)
\(x=32:\left(-8\right)\)
\(x=-4\)
\(c,\left(-9\right)x+3=\left(-2\right)\left(-7\right)+16\)
\(\left(-9\right)x+3=14+16\)
\(\left(-9\right)x+3=30\)
\(\left(-9\right)x=30-3\)
\(\left(-9\right)x=27\)
\(x=27:\left(-9\right)\)
\(x=-3\)
\(a.5x=-20\\ x=-20:5=-4\\ b.\dfrac{5}{6}-x=\dfrac{7}{4}\\ x=\dfrac{5}{6}-\dfrac{7}{4}=\dfrac{-11}{12}\)
\(5x=-20\Leftrightarrow x=-20:5\Leftrightarrow x=-4\)
\(\dfrac{5}{6}-x=\dfrac{7}{4}\Leftrightarrow x=\dfrac{5}{6}-\dfrac{7}{4}\Leftrightarrow x=-\dfrac{11}{12}\)
a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
=> 6(2x-7) = 9(x-3)
=> 12x - 42 = 9x - 27
=> 12x - 9x = -27 + 42
=> 3x = 15
=> x = 5
Vậy x = 5
b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
=> -7(x + 27) = 6(x + 1)
=> -7x - 189 = 6x + 6
=> -7x - 6x = 6 + 189
=> -13x = 195
=> x = -15
Vậy x = -15
a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)
\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)
\(\Leftrightarrow12x-42=9x-27\)
\(\Leftrightarrow12x-9x=-27+42\)
\(\Leftrightarrow3x=15\)
hay x=5
Vậy: x=5
b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)
\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)
\(\Leftrightarrow6x+6=-7x+189\)
\(\Leftrightarrow6x+7x=189-6\)
\(\Leftrightarrow13x=183\)
hay \(x=\dfrac{183}{13}\)
Vậy: \(x=\dfrac{183}{13}\)
a) \(\dfrac{2}{5}\cdot x=\dfrac{1}{2}+\dfrac{6}{8}\)
\(\dfrac{2}{5}\cdot x=\dfrac{5}{4}\)
\(x=\dfrac{5}{4}\div\dfrac{2}{5}\)
\(x=\dfrac{25}{8}\)
b) \(\dfrac{20}{7}-x=\dfrac{19}{7}\div\dfrac{3}{2}\)
\(\dfrac{20}{7}-x=\dfrac{19}{7}\cdot\dfrac{2}{3}\)
\(\dfrac{20}{7}-x=\dfrac{38}{21}\)
\(x=\dfrac{20}{7}-\dfrac{28}{21}\)
\(x=\dfrac{22}{21}\)
b) \(3^x\cdot3^2+3^x=7290\)
\(3^x\left(3^2+1\right)=720\)
\(3^x\cdot10=7290\)
\(=>3^x=729=3^6\)
=> \(x=6\)