Toán Lớp 8: Tìm x biết: a/ 3x( 2x +7) – ( x+ 1)( 6x -5) = 10 b/ (x – 4) ² – 36 =0 c/3(2x – 1) – 5( x – 3) =4

Question

Toán Lớp 8:  Tìm x biết: a/ 3x( 2x +7) – ( x+ 1)( 6x -5) = 10                b/ (x – 4) ² – 36 =0 c/3(2x – 1) – 5( x – 3) =4                              d/ (2x – 1) ² – (x+3) ² = 0 e/ (x – 3)(x – 7) – x ²  = 1                              f/ ( x +2) ² – ( x – 2)( x-1) = 0 g/ ( x – 1) ² – ( x – 2)( x+2)  = 0                      h/ ( 10x + 9).x – ( 5x – 1)(2x +3) = 8

kimnguenoanh · Answer

Giải đáp + Lời giải và giải thích chi tiết ! to Tìm x: a) 3x(2x+7)-(x+1)(6x-5) = 10 <=> 6x^2+21x-6x^2+5x-6x+5 = 10 <=> (6x^2-6x^2)+(21x+5x-6x)+5 = 10 <=> 20x = 5 <=> x = 1/4 Vậy S= {1/4} b) (x-4)^2-36 = 0 <=> (x-4)^2-6^2 = 0 <=> (x-4-6)(x-4+6) = 0 <=> (x-10)(x+2) = 0 <=> ( left[ begin{array}{l}x-10=0 x+2=0 end{array} right. ) <=> ( left[ begin{array}{l}x=10 x=-2 end{array} right. ) Vậy S= {10; -2} Áp dụng: a^2-b^2 = (a-b)(a+b) c) 3(2x-1)-5(x-3) = 4 <=> 6x-3-5x+15 = 4 <=> (6x-5x)+(-3+15) = 4 <=> x+12 = 4 <=> x = -8 Vậy S= {-8} d) (2x-1)^2-(x+3)^2 = 0 <=> (2x-1-x-3)(2x-1+x+3) = 0 <=> (x-4)(3x+2) = 0 <=> ( left[ begin{array}{l}x-4=0 3x+2=0 end{array} right. ) <=> ( left[ begin{array}{l}x=4 3x=-2 end{array} right. ) <=> ( left[ begin{array}{l}x=4 x=- dfrac{2}{3} end{array} right. ) Vậy S= {4; -(2)/3} Áp dụng: a^2-b^2 = (a-b)(a+b) e) (x-3)(x-7)-x^2 = 1 <=> x^2-7x-3x+21-x^2 = 1 <=> (x^2-x^2)+(-7x-3x)+21 = 1 <=> -10x = -20 <=> x = 2 Vậy S= {2} f) (x+2)^2-(x-2)(x-1) = 0 <=> x^2+4x+4-x^2+x+2x-2 = 0 <=> (x^2-x^2)+(4x+x+2x)+(4-2) = 0 <=> 7x+2 = 0 <=> 7x = -2 <=> x = -(2)/7 Vậy S= {-(2)/7} g) (x-1)^2-(x-2)(x+2) = 0 <=> x^2-2x+1-x^2+4 = 0 <=> (x^2-x^2)-2x+(1+4) = 0 <=> -2x+5 = 0 <=> -2x = -5 <=> x = 5/2 Vậy S= {5/2} Áp dụng: +) (a-b)^2 = a^2-2ab+b^2 +) a^2-b^2 = (a-b)(a+b) h) (10x+9). x-(5x-1)(2x+3) = 8 <=> 10x^2+9x-10x^2-15x+2x+3 = 8 <=> (10x^2-10x^2)+(9x-15x+2x)+3 = 8 <=> -4x = 5 <=> x = -(5)/4 Vậy S= {-(5)/4}

maianh67 · Answer

@Sun        Bạn xem h&igrave;nh