giải phương trình: `(3-x)^4+(2-x)^4=(5-2x)^4` `(x-7)^4+(x-8)^4=(15-2x)^4` `(x+1)^3+(x-2)^3=(2x-1)^3`

Đáp án + Giải thích các bước giải:

`(3-x)^4+(2-x)^4=(5-2x)^4(1)`

Đặt `a=3-x;b=2-x=>a+b=3-x+2-x=5-2x`. Pt `(1)` trở thành:

`a^4+b^4=(a+b)^4`

`<=>a^4+b^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4`

`<=>2ab(2a^2+3ab+2b^2)=0`

TH1: `2ab=0`

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

`<=>[(3-x=0),(2-x=0):}`

`<=>[(x=3),(x=2):}`

TH2: `2a^2+3ab+2b^2=0`

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

`<=>2(x^2-6x+9)+3(x^2-5x+6)+2(x^2-4x+4)=0`

`<=>2x^2-12x+18+3x^2-15x+18+2x^2-8x+8=0`

`<=>7x^2-35x+44=0`

`<=>7(x^2-5x+44/7)=0`

`<=>(x^2-5x+25/4)+1/28=0`

`<=>(x-5/2)^2=-1/28` (vô lí)

Vậy `S={3;2}`

`----`

`(x-7)^4+(x-8)^4=(15-2x)^4(1)`

Đặt `a=x-7;b=x-8=>-(a+b)=-(x-7+x-8)=-(2x-15)=15-2x`. Pt `(1)` trở thành:

`a^4+b^4=[-(a+b)]^4`

`<=>a^4+b^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4`

`<=>2ab(2a^2+3ab+2b^2)=0`

TH1: `2ab=0`

`<=>2.(x-7)(x-8)=0`

`<=>[(x-7=0),(x-8=0):}`

`<=>[(x=7),(x=8):}`

TH2: `2a^2+3ab+2b^2=0`

`<=>2.(x-7)^2+3.(x-7)(x-8)+2.(x-8)^2=0`

`<=>2(x^2-14x+49)+3(x^2-15x+56)+2(x^2-16x+64)=0`

`<=>2x^2-28x+98+3x^2-45x+168+2x^2-32x+128=0`

`<=>7x^2-105x+394=0`

`<=>7(x^2-15x+394/7)=0`

`<=>(x^2-15x+225/4)+1/28=0`

`<=>(x-15/2)^2=-1/28` (vô lí)

Vậy `S={7;8}`

`----`

`(x+1)^3+(x-2)^3=(2x-1)^3(1)`

Đặt `a=x+1;b=x-2=>a+b=x+1+x-2=2x-1`. Pt `(1)` trở thành:

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

`<=>a^3+b^3=a^3+3a^2b+3ab^2+b^3`

`<=>3ab(a+b)=0`

TH1: `3ab=0`

`<=>3.(x+1)(x-2)=0`

`<=>[(x+1=0),(x-2=0):}`

`<=>[(x=-1),(x=2):}`

TH2: `a+b=0`

`<=>x+1+x-2=0`

`<=>2x-1=0`

`<=>2x=1`

`<=>x=1/2`

Vậy `S={-1;2;1/2}`