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\(x^3:\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow x^3:\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow x^3=\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(\Rightarrow x^3=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow x=\dfrac{1}{2}\)
\(x^3:\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\Rightarrow x^3=\dfrac{1}{2}.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}.\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow x=\dfrac{1}{2}\)
Đề yêu cầu gì vậy em, phải có yêu cầu cụ thể mới cứu em được chứ nhỉ?
\(x+-\dfrac{1}{9}=-\dfrac{1}{6}\)
\(\Rightarrow x-\dfrac{1}{9}=-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{1}{6}+\dfrac{1}{9}\)
\(\Rightarrow x=-\dfrac{1}{18}\)
_______________
\(5x-3x-\dfrac{5}{6}=\dfrac{2}{3}\)
\(\Rightarrow2x-\dfrac{5}{6}=\dfrac{2}{3}\)
\(\Rightarrow2x=\dfrac{2}{3}+\dfrac{5}{6}\)
\(\Rightarrow2x=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}:2\)
\(\Rightarrow x=\dfrac{3}{4}\)
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
\(\begin{array}{l}a)3{x^7}:\dfrac{1}{2}{x^4} = (3:\dfrac{1}{2}).({x^7}:{x^4}) = 6{x^3}\\b)( - 2x):x = [( - 2):1].(x:x) = - 2\\c)0,25{x^5}:( - 5{x^2}) = [0,25:( - 5)].({x^5}:{x^2}) = - 0,05.{x^3}\end{array}\)
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
