Toán Lớp 8: Tìm x a) (x-2)²-(x-3).(x+3)=5 b) x²+5x+6=0

Question

Toán Lớp 8:  Tìm x a) (x-2)²-(x-3).(x+3)=5 b) x²+5x+6=0

thucquyenlan · Answer

Giải đáp:     a, (x - 2)^2 - (x - 3).(x + 3) = 5   &harr; x^2 - 4x + 4 - (x^2 - 3^2) = 5   &harr;  x^2 - 4x + 4 - (x^2 - 9) = 5   &harr;  x^2 - 4x + 4 - x^2 + 9 = 5   &harr; (x^2 - x^2) -4x + (4 + 9) = 5   &harr; -4x + 13 = 5   &harr; -4x = 5 - 13   &harr; -4x = -8   &harr; x = (-8) : (-4)   &harr; x = 2   Vậy x = 2   -------------------------   b, x^2 + 5x + 6 = 0   &harr; x^2 + 2x + 3x + 6 = 0   &harr; x.(x + 2) + 3.(x + 2) = 0   &harr; (x + 2).(x + 3 ) =0   &harr; $ left[ begin{matrix} x+2 = 0   x + 3 = 0 end{matrix} right.$   &harr; $ left[ begin{matrix} x = -2   x  = -3 end{matrix} right.$   Vậy x &isin; {-2; -3}   #dariana

lanthuhoa · Answer

Giải đáp: a) (x-2)²-(x-3).(x+3)=5 <=> ( x^2 - 2.x.2 + 2^2) - ( x^2 - 3^2) - 5 = 0 <=> x^2 - 4x +4 - x^2 + 9 - 5 = 0 <=> ( x^2 - x^2) - 4x + ( 4 + 9 -5) = 0 <=> - 4x + 8 = 0 <=> -4x = -8 <=> x = 2 Vậy S={ 2} $ $ b) x²+5x+6=0 <=> x^2 + 3x + 2x +6 = 0 <=> x( x +3) + 2( x +3) = 0 <=> ( x +2)( x +3) = 0 <=> ( left[ begin{array}{l}x+2 = 0 x+ 3 = 0 end{array} right. ) $ $ <=> ( left[ begin{array}{l}x=-2 x=-3 end{array} right. ) Vậy S={ -2 ; -3} Áp dụng: 1. ( A - B)^2 = A^2 - 2AB + B^2 2. A^2 - B^2 = ( A - B)( A + B) 3. A . B = 0 <=> ( left[ begin{array}{l}A=0 B=0 end{array} right. ) 4. Phân tích thành nhân tử.