Toán Lớp 8: Bài1:tìm x a) (5x+1) ² -(5x-3) (5x+3)=20 b) (2x+3) ²-(2x+1) (2x-1)=22 c) (5x-2) ²+(2-5x) (3x+1)=0 Mn giúp mk đc hok ????????

Question

Toán Lớp 8:  Bài1:tìm x a) (5x+1) ² -(5x-3) (5x+3)=20 b) (2x+3) ²-(2x+1) (2x-1)=22 c) (5x-2) ²+(2-5x) (3x+1)=0 Mn giúp mk đc hok ????????

maingocquynhnhu2k1 · Answer

Giải đáp: + Lời giải và giải thích chi tiết:    a)   $(5x + 1)&sup2; - (5x - 3)(5x + 3) = 20$   $25x&sup2; + 10x + 1 - 25x&sup2; + 9 = 20$   $10x + 10 = 20$   $10x = 10$   $x =1 $   b)    $ (2x+3) &sup2;-(2x+1) (2x-1)=22$     $4x&sup2; + 12x + 9 - 4x&sup2; + 1 = 22$     $12x + 10 = 22$     $12x = 12$     $x = 1$     c)     $(5x-2) &sup2;+(2-5x) (3x+1)=0$     $25x&sup2; - 20x + 4 + (6x + 2 - 15x&sup2; - 5x) = 0$     $25x&sup2; - 20x + 4 - 15x&sup2; + x + 2 = 0$     $10x&sup2; - 19x + 6 = 0$     $10x&sup2; - 4x - 15x + 6 = 0$     $2x(5x - 2) - 3(5x - 2) = 0$     $(2x - 3)(5x - 2) = 0$      ( left[  begin{array}{l}x= dfrac{3}{2}  x= dfrac{2}{5} end{array}  right. )

giahan44 · Answer

Giải đáp:   a)x=1   b)x=1   c)x&isin;{2/5;3/2}   Lời giải và giải thích chi tiết:   a)(5x+1)&sup2;-(5x-3)(5x+3)=20   &hArr;25x&sup2;+10x+1-[(5x)&sup2;-3&sup2;]=20   &hArr;25x&sup2;+10x+1-(25x&sup2;-9)=20   &hArr;25x&sup2;+10x+1-25x&sup2;+9=20   &hArr;(25x&sup2;-25x&sup2;)+10x+(1+9)=20   &hArr;10x+10=20   &hArr;10x=20-10   &hArr;10x=10   &hArr;x=10:10   &hArr;x=1   Vậy x=1   b)(2x+3)&sup2;-(2x+1)(2x-1)=22   &hArr;4x&sup2;+12x+9-[(2x)&sup2;-1&sup2;]=22   &hArr;4x&sup2;+12x+9-(4x&sup2;-1)=22   &hArr;4x&sup2;+12x+9-4x&sup2;+1=22   &hArr;(4x&sup2;-4x&sup2;)+12x+(9+1)=22   &hArr;12x+10=22   &hArr;12x=22-10   &hArr;12x=12   &hArr;x=12:12   &hArr;x=1   Vậy x=1   c)(5x-2)&sup2;+(2-5x)(3x+1)=0   &hArr;(2-5x)&sup2;+(2-5x)(3x+1)=0   &hArr;(2-5x)(2-5x+3x+1)=0   &hArr;(2-5x)(3-2x)=0   &hArr;$ left[ begin{matrix} 2-5x=0   3-2x=0 end{matrix} right.$   &hArr;$ left[ begin{matrix} 5x=2   2x=3 end{matrix} right.$   &hArr;$ left[ begin{matrix} x= dfrac{2}{5}   x= dfrac{3}{2} end{matrix} right.$   Vậy x&isin;{2/5;3/2}