24+5-x+6=3 tìm x
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khong them viet ct :v
(x + 3)(x + 4)(x + 5)(x + 6) = 24
x^4 + 6x^3 + 5x^3 + 30x^2 + 4x^3 + 24x^2 + 20x^2 + 120x + 3x^2 + 18x^2 + 15x^2 + 90x + 12x^2 + 72x + 60x + 360 = 24
x^4 + 18x^3 + 119x^2 + 342x + 360 = 24
x^4 + 18x^3 + 119x^2 + 342x + 360 - 24 = 0
x^4 + 18x^3 + 119x^2 + 342x + 336 = 0
(x^3 + 16x^2 + 87x + 168)(x + 2) = 0
(x^2 + 9x + 24)(x + 7)(x + 2) = 0
vi x^2 + 9x + 24 # 0 nen:
=> x + 7 = 0 => x = -7
x + 2 = 0 => x = -2
=> x = -7; -2



\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)
\(\dfrac{1}{6}+x=\dfrac{14}{5}\)
\(=>x=\dfrac{79}{30}\)
\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)
\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)
\(\dfrac{32}{35}:x=\dfrac{11}{6}\)
\(x=\dfrac{192}{385}\)
\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)
\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)
\(x:\dfrac{3}{4}=\dfrac{38}{30}\)
\(=>x=\dfrac{19}{20}\)
\(a,\dfrac{1}{2}:3+x=\dfrac{14}{5}\\ \Leftrightarrow x+\dfrac{1}{6}=\dfrac{14}{5}\\ \Leftrightarrow x=\dfrac{79}{30}\\ b,\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\\ \Leftrightarrow x=\dfrac{192}{385}\\ c,\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\\ \Leftrightarrow\dfrac{4}{3}x=\dfrac{19}{15}\\ \Leftrightarrow x=\dfrac{19}{20}\)

a) ? × 4 = 28
28 : 4 = 7
Vậy 7 × 4 = 28.
b) ? × 3 = 12
12 : 3 = 4
Vậy 4 × 3 = 12.
c) 6 × ? = 24
24 : 6 = 4
Vậy 6 × 4 = 24.

\(\left(\frac{-5}{3}\right)^3\)< x < \(\frac{-24}{35}.\frac{-5}{6}\)
\(\Rightarrow\frac{-125}{27}\)< x < \(\frac{4}{7}\)
\(\Rightarrow-4,\left(629\right)\)< x < \(0,\left(571428\right)\)ư
Vậy x = -4; -3; -2; -1; 0
24+5-x+6=3
24+5-x=3-6
24+5-x=-3
29-x=-3
x=29 - -3
x=32
35 - x = 3
x = 35 - 3
x = 32