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a) \(\begin{array}{l}(4x - 3)(x + 2) = 4x(x + 2) - 3(x + 2)\\ = 4{x^2} + 8x - 3x - 6\end{array}\)
\( = 4{x^2} + 5x - 6\)
b) \((5x + 2)( - {x^2} + 3x + 1)\)
\( = 5x( - {x^2} + 3x + 1) + 2( - {x^2} + 3x + 1)\)
\( = - 5{x^3} + 15{x^2} + 5x - 2{x^2} + 6x + 2\)
\( = - 5{x^3} + 13{x^2} + 11x + 2\)
c) \((2{x^2} - 7x + 4)( - 3{x^2} + 6x + 5)\)
\( = 2{x^2}( - 3{x^2} + 6x + 5) - 7x( - 3{x^2} + 6x + 5) + 4( - 3{x^2} + 6x + 5)\)
\( = 2{x^2}( - 3{x^2}) + 2{x^2}.6x + 2{x^2}.5 + 7x.3{x^2} - 7x.6x - 7x.5 + 4( - 3{x^2}) + 4.6x + 4.5\)
\(= - 6{x^4} + 33{x^3} - 44{x^2} - 11x + 20\)
\(a.=\left(\frac{83}{5}-\frac{68}{5}\right).-\frac{1}{3}+\frac{3}{4}\)
\(=\frac{15}{5}.-\frac{1}{3}+\frac{3}{4}\)
\(=3.-\frac{1}{3}+\frac{3}{4}\)
\(=-1+\frac{3}{4}\)
\(=-\frac{1}{4}\)
a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
\(\begin{array}{l}a)3{x^7}:\dfrac{1}{2}{x^4} = (3:\dfrac{1}{2}).({x^7}:{x^4}) = 6{x^3}\\b)( - 2x):x = [( - 2):1].(x:x) = - 2\\c)0,25{x^5}:( - 5{x^2}) = [0,25:( - 5)].({x^5}:{x^2}) = - 0,05.{x^3}\end{array}\)
Tham khảo:
a) \((4{x^2} - 5):(x - 2) = \dfrac{{4{x^2} - 5}}{{x - 2}} = 4x + 8 + \dfrac{{11}}{{x - 2}}\)
Vậy \( (4{x^2} - 5):(x - 2)= 4x + 8 + \dfrac{{11}}{{x - 2}}\)
b) \((3{x^3} - 7x + 2):(2{x^2} - 3) = \dfrac{{3{x^3} - 7x + 2}}{{2{x^2} - 3}}\)
Vậy \( (3{x^3} - 7x + 2):(2{x^2} - 3)= \dfrac{3}{2}x + \dfrac{{\dfrac{-5}{2}x + 2}}{{2{x^2} - 3}}\)