Toán Lớp 8: Bài 1: Tính a) (x+2)^2-(x-3)(x+3) Bài 2: Tìm x a) 5x(x-1)+2-2x=0

Question

Toán Lớp 8:  Bài 1: Tính a) (x+2)^2-(x-3)(x+3) Bài 2: Tìm x a) 5x(x-1)+2-2x=0

hatuanlan · Answer

Giải đáp:    $a) (x+2)&sup2;-(x-3).(x+3)$   $= x&sup2;+4x+4-x&sup2;+9$   $= (x&sup2;-x&sup2;)+4x+(4+9)$   $= 4x+13$   $b) 5x.(x-1)+2-2x=0$   $&hArr; 5x.(x-1)-2.(x-1)=0$   $&hArr; (5x-2).(x-1)=0$   $&hArr;$  ( left[  begin{array}{l}5x-2=0  x=1 end{array}  right. ) $&hArr;$   ( left[  begin{array}{l}5x=2  x=1 end{array}  right. ) $&hArr;$   ( left[  begin{array}{l}x= dfrac{2}{5}  x=1 end{array}  right. )    $Vậy$ $x= dfrac{2}{5}$ $hoặc$ $x=1$

ngoclandiep · Answer

Bài 1: a) (x+2)^2-(x-3)(x+3) = (x+2)^2-(x-3)(x+3) = x^2+4x+4-(x-3)(x+3) = x^2 + 4x+4-(x^2-9) = x^2+4x+4-x^2+9 = 4x+4 + 9 = 4x+13 Bài 2: a) 5x(x-1)+2-2x=0 <=> 5x^2-5x+2-2x=0 <=> 5x^2-7x+2=0 <=> (5x^2-5x)+(-2x+2)=0 <=> 5x(x-1)-2(x-1)=0 <=> (x-1)(5x-2)=0 <=> ( left[ begin{array}{l}x-1=0 5x-2=0 end{array} right. ) <=> ( left[ begin{array}{l}x=1 x= dfrac{2}{5} end{array} right. ) Vậy x in {1;2/5}