Toán Lớp 8: Chứng minh đẳng thức :
[$\frac{2}{3x}$.($\frac{x+1}{3x}$- x – 1)] : $\frac{x-1}{x}$ = $\frac{2x}{x-1}$
Giúp mình với ạ, mình cảm ơn !
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Toán Lớp 8: Chứng minh đẳng thức :
[$\frac{2}{3x}$.($\frac{x+1}{3x}$- x – 1)] : $\frac{x-1}{x}$ = $\frac{2x}{x-1}$
Giúp mình với ạ, mình cảm ơn !
Comments ( 1 )
VT = \left[ {\dfrac{2}{{3x}}.\left( {\dfrac{{x + 1 – 3{x^2} – 3x}}{{3x}}} \right)} \right].\dfrac{x}{{x – 1}}\\
= \dfrac{2}{{3x}}.\dfrac{{ – 3{x^2} – 2x + 1}}{{3x}}.\dfrac{x}{{x – 1}}\\
= \dfrac{2}{{3x}}.\dfrac{{\left( {1 – 3x} \right)\left( {x + 1} \right)}}{{3x}}.\dfrac{x}{{x – 1}}\\
= \dfrac{{2\left( {x + 1} \right)\left( {1 – 3x} \right)}}{{9x\left( {x – 1} \right)}} \ne VP
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