`\color{#7D26CD}{q}\color{#8968CD}{u}\color{#9370DB}{a}\color{#9F79EE}{c}\color{#AB82FF}{k}`

Bài `13:`

`(x^2+9)(x^2+9x) =22(x-1)^2`

`x^4 +9x^3 +9x^2 +81x=22(x^2-2x+1) `

`x^4+9x^3+9x^2+81x=22x^2 -44x+22`

`x^4+9x^3-13x^2 +125x-22=0`

`(x^4 -2x^3 +11x^2) +(11x^3 -22x^2 +121x) -(2x^2 -4x+22) =0`

`x^2(x^2-2x+11) +11x(x^2-2x+11) -2(x^2-2x+11) =0`

`(x^2-2x+11)(x^2+11x-2) =0`

`TH1: x^2-2x+11=0`

`x^2-2x+1+10=0`

`(x-1)^2 +10=0`

Vì `(x-1)^2>=0AAx =>(x-1)^2+10>=10>0AAx`

`=>` phương trình vô nghiệm

`TH2: x^2+11x-2=0`

`x^2 +2x*11/2 +(11/2)^2 -(11/2)^2 -2=0`

`(x+11/2)^2 -121/4 -2=0`

`(x+11/2)^2 -129/4=0`

`(x+11/2)^2 -((\sqrt129)/2)^2 =0`

`(x+11/2 -(\sqrt129)/2)(x+11/2 +(\sqrt129)/2) =0`

`(x+(11-sqrt129)/2)(x+(11+sqrt129)/2) =0`

`TH1: x+(11-sqrt129)/2=0`

`=>x=(11-sqrt129)/2`

`TH2: x +(11+sqrt129)/2=0`

`=>x=(11+sqrt129)/2`

Vậy phương trình có nghiệm `x in {(-11+\sqrt(129))/2;(-11-\sqrt(129))/2}`

Bài `14:`

`(x^2-9)^2 =12x+1`

`x^4 -18x^2 +81=12x+1`

`x^4-18x^2 -12x+80=0`

`(x^4 +6x^3 +10x^2) -(6x^3 +36x^2+60x) +(8x^2 +48x+80) =0`

`x^2(x^2+6x+10) -6x(x^2+6x+10) +8(x^2+6x+10) =0`

`(x^2-6x+8)(x^2+6x+10) =0`

`(x^2-2x-4x+8)(x^2+6x+10) =0`

`[x(x-2) -4(x-2)](x^2+6x+10) =0`

`(x-2)(x-4)(x^2+6x+10) =0`

`TH1:x-2=0=>x=2`

`TH2:x-4=0=>x=4`

`TH3:x^2+6x+10=0`

`x^2+6x+9+1=0`

`(x+3)^2+1=0`

Vì `(x+3)^2>=0AAx =>(x+3)^2+1>=1>0AAx`

`=>` phương trình vô nghiệm

Vậy phương trình có nghiệm `x in {2;4}`