Question

Mí bạn giúp mik chứng minh bài này nhé!:)))))))))))))))

\(\left(\dfrac{x\sqrt{x}-3\sqrt{3x}}{x-27}+\dfrac{x^3-x^2+x}{3\sqrt{3x}+x\sqrt{x}}\right):\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\) =1

Mí bạn giúp mik giải bài này trong hôm nay hoặc ngày mai nhé!:)))))))))))))))))))))Cảm ơn nhiều!:)))))))))))))))))))))))))))))

Answer 1

Chứng minh:\(\left(\dfrac{x\sqrt{x}-3\sqrt{3x}}{x-27}+\dfrac{x^3-x^2+x}{3\sqrt{3x}+x\sqrt{x}}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)= 1

Biến đổi vế trái ta được:

VT=\(\left(\dfrac{(x\sqrt{x}-3\sqrt{3x)}\times\left(3\sqrt{3x}+x\sqrt{x}\right)}{(x-27)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}+\dfrac{\left(x-27\right)\times(x^3-x^2+x)}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)

=\(\left(\dfrac{x^3-27x^2}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}+\dfrac{\left(x-27\right)\times\left(x^3-x^2+x\right)}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)

=\(\left(\dfrac{x^2}{3\sqrt{3x}+x\sqrt{x}}+\dfrac{x^3-x^2+x}{3\sqrt{3x}+x\sqrt{x}}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)

=\(\dfrac{x^3+x}{3\sqrt{3x}+x\sqrt{x}}\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)

=\(\dfrac{(x^3+x)\times\left(\sqrt{x}+3\sqrt{3}\right)}{(3\sqrt{3x}+x\sqrt{x})\times(x^2+1)}\)

=\(\dfrac{x\times\left(x^2+1\right)\times\left(\sqrt{x}+3\sqrt{3}\right)}{\left(x^2+1\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\)

=\(\dfrac{x\times\left(\sqrt{x}+3\sqrt{3}\right)}{\sqrt{x}\times\left(x+3\sqrt{3}\right)}\)

=\(\dfrac{x\times\left(\sqrt{x}+3\sqrt{3}\right)}{x\times\left(\sqrt{x}+3\sqrt{3}\right)}\)= 1 =VP

Vậy đẳng thức được chứng minh