Toán Lớp 8: tìm x a 5x(x-3)(x+3)-(2x-3)^2-5(x+2)^3+34(x+2)=1 b) (x-2)^3+ 6(x+1)^2 - (x-3)(x^2+3x+9)=97 -

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Toán Lớp 8: tìm x a 5x(x-3)(x+3)-(2x-3)^2-5(x+2)^3+34(x+2)=1 b) (x-2)^3+ 6(x+1)^2 – (x-3)(x^2+3x+9)=97

Khánh Giao Tháng Hai 21, 2023 Toán học 1 Comment 0 views

Toán Lớp 8: tìm x a 5x(x-3)(x+3)-(2x-3)^2-5(x+2)^3+34(x+2)=1
b) (x-2)^3+ 6(x+1)^2 – (x-3)(x^2+3x+9)=97

Comments ( 1 )

trucoanh55
21 Tháng Hai, 2023 at 1:31 chiều

Reply

a) 5x(x-3)(x+3)-(2x-3)^2-5(x+2)^3+34(x+2)=1

<=> 5x(x^2-9)-(4x^2-12x+9)-5(x^2+4x+4)+34x+68=1

<=> 5x^3-45x-4x^2+12x-9-5x^2-20x-20+34x+67=0

<=> 5x^3-9x^2-19x+38=0

<=> 5x^3+10x^2-19x^2-38x+19x+38=0

<=> 5x^2(x+2)-19x(x+2)+19(x+2)=0

<=> (x+2)(5x^2-19x+19)=0

Do 5x^2-19x+19=5(x^2-19/5x+19/5)=5(x-19/10)^2+19/20>0 với AAx

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

Vậy S={-2}

b) (x-2)^3+6(x+1)^2-(x-3)(x^2+3x+9)=97

<=> x^3-6x^2+12x-8+6(x^2+2x+1)-(x^3-27)-97=0

<=> x^3-6x^2+12x-105+6x^2+12x+6-x^3+27=0

<=> 24x-72=0

<=> x-3=0

<=> x=3

Vậy S={3}

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