1+1/64x^3=....^3+(...)^3=(...+...)(....-...+....)

Question

cattuong · Answer

`Hđt` thứ `6:a^3+b^3=(a+b)(a^2-ab+b^2)`   `&rArr;1+1/64x^3=1^3+(1/4x)^3=(1+1/4x)[1^2-1.1/4x+(1/4x)^2]`   `=(1+1/4x)(1-1/4x+1/16x^2)`

maianhnhi · Answer

Đ&aacute;p &aacute;n:        `1+ 1/(64) x&sup3;`   `= 1&sup3; +1/(4^3) x&sup3;`   `= 1&sup3; + (1/4  x)^3`   `= (1 +1/4 x)( 1- 1/4 x +1/(16) x&sup2;)`    _____________________      `1+ 1/(64x&sup3;)`   `= 1&sup3; + (1/(4x))^3`   `= (1+ 1/(4x))( 1- 1/(4x) +1/(16x&sup2;))`   AD: `a&sup3; +b&sup3; = (a+b)(a&sup2;-ab+b&sup2;)`

1+1/64x^3=….^3+(…)^3=(…+…)(….-…+….)

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