27(x-3)^3 = 8(x-2)^3+ +(x-5)^3 HỨA VOTE ĐẦY ĐỦ Ạ

Đáp án:

`x \in {2; 5; 3}`

Giải thích các bước giải:

`27(x-3)^3 = 8(x-2)^3+(x-5)^3`

`<=> 27(x^3-9x^2+27x-27) = 8(x^3-6x^2+12x-8)+(x^3-15x^2+75x-125)`

`<=> 27x^3-243x^2+729x-729 = 8x^3-48x^2+96x-64+x^3-15x^2+75x-125`

`<=> 27x^3-243x^2+729x-729-8x^3+48x^2-96x+64-x^3+15x^2-75x+125 = 0`

`<=> (27x^3-8x^3-x^3)+(-243x^2+48x^2+15x^2)+(729x-96x-75x)+(-729+64+125) = 0`

`<=> 18x^3-180x^2+558x-540 = 0`

`<=> 18(x^3-10x^2+31x-30) = 0`

`<=> x^3-10x^2+31x-30 = 0`

`<=> x^3-3x^2-7x^2+21x+10x-30 = 0`

`<=> (x^3-3x^2)-(7x^2-21x)+(10x-30) = 0`

`<=> x^2(x-3)-7x(x-3)+10(x-3) = 0`

`<=> (x^2-7x+10)(x-3) = 0`

`<=> (x^2-5x-2x+10)(x-3) = 0`

`<=> [(x^2-5x)-(2x-10)](x-3) = 0`

`<=> [x(x-5)-2(x-5)](x-3) = 0`

`<=> (x-2)(x-5)(x-3) = 0`

`<=>` \(\left[ \begin{array}{l}x-2=0\\x-5=0\\x-3=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2\\x=5\\x=3\end{array} \right.\) 

Vậy `x \in {2; 5; 3}`

Áp dụng:

`+)` `(a+b)^3 = a^3+3a^2b+3ab^2+b^3`

`+)` `(a-b)^3 = a^3-3a^2b+3ab^2-b^3`