Toán Lớp 8: a) 2(x-1)^2+(x+3)^2=3(x-2)(x+1) b) (4x^3+3x+1)(4x^3-3x+1)-(4x^3+1)^2=36 Giúp mk vs

Question

Toán Lớp 8:  a) 2(x-1)^2+(x+3)^2=3(x-2)(x+1) b) (4x^3+3x+1)(4x^3-3x+1)-(4x^3+1)^2=36 Giúp mk vs

tuyethutanhthu · Answer

Giải đáp:       Lời giải và giải thích chi tiết:   a)   2(x-1)^2 + (x+3)^2 = 3(x-2)(x+1)   -> 3x^2 + 2x + 11 = 3x^2 - 3x - 6   -> 3x^2 + 2x = 3x^2 - 3x - 17   -> 2x = -3x - 11   -> 5x = -17   -> x = -17/5   Vậy x  in {-17/5}   b)   (4x^3+3x+1)(4x^3-3x+1)-(4x^3+1)^2 = 36   -> 16x^6 + 8x^3 - 9x^2 + 1 - 16x^6 - 8x^3 - 1 = 36   -> (16x^6-16x^6) + (8x^3-8x^3) + (1-1) - 9x^2 = 36   -> -9x^2 = 36   -> x^2 = -4   -> x &isin; &empty;   Vậy x &isin; &empty;

maingocquynhnhu2k1 · Answer

$ begin{array}{l} a)2{ left( {x - 1}  right)^2} + { left( {x + 3}  right)^2} = 3 left( {x - 2}  right) left( {x + 1}  right)     Leftrightarrow 2{x^2} - 4x + 2 + {x^2} + 6x + 9 = 3 left( {{x^2} - x - 2}  right)     Leftrightarrow 3{x^2} + 2x + 11 = 3{x^2} - 3x - 6     Leftrightarrow 5x =  - 5  Leftrightarrow x =  - 1   b)(4{x^3} + 3x + 1)(4{x^3} - 3x + 1) - {(4{x^3} + 1)^2} = 36     Leftrightarrow { left( {4{x^3} + 1}  right)^2} - { left( {3x}  right)^2} - { left( {4{x^3} + 1}  right)^2} = 36     Leftrightarrow  - 9{x^2} = 36     Leftrightarrow {x^2} =  - 4     Rightarrow x  in  emptyset   end{array}$