Toán Lớp 8: Tìm x biết a.x^2-25-(x+5)=0 b.(10x+9)x-(5x-1)(2x+3)=0 c.(2x+3)(x-1)+(2x-3)(1-x)=0 d. (5x-4)^2-49x^2=0 e.x^2+3x-10=0

Question

Toán Lớp 8:  Tìm x biết  a.x^2-25-(x+5)=0 b.(10x+9)x-(5x-1)(2x+3)=0   c.(2x+3)(x-1)+(2x-3)(1-x)=0 d. (5x-4)^2-49x^2=0 e.x^2+3x-10=0

quynhmaithu · Answer

Giải đáp: + Lời giải và giải thích chi tiết:   $a$) $x^2 - 25 - (x+5) = 0$   $&hArr; x^2 - 25 - (x+5) = 0$   $&hArr; (x-5)(x+5) - (x+5) = 0$   $&hArr; (x+5)(x-6) = 0$   $&rArr;$  ( left[  begin{array}{l}x=6  x=-5 end{array}  right. )      Vậy $x$ $&isin;$ {-5;6}.    $b$) $(10x+9)x - (5x-1)(2x+3) = 0$   $&hArr; 10x^2 + 9x - (10x^2 - 2x + 15x - 3) = 0$   $&hArr; 10x^2 + 9x - 10x^2 + 2x - 15x + 3= 0$   $&hArr; -4x + 3 =0$   $&hArr; x =  dfrac{3}{4}$     Vậy x=3/4.   $c$) $(2x+3)(x-1) + (2x-3)(1-x) = 0$   $&hArr; (x-1)(2x+3-2x+3)=0$   $&hArr; 6(x-1) = 0$   $&hArr; x=1$      Vậy $x=1$.   $d$) $(5x-4)^2 - 49x^2 = 0$   $&hArr; (5x-4)^2 - (7x)^2 = 0$   $&hArr; (5x-4+7x)(5x-4-7x) =0$   $&hArr; (12x - 4)(-2x-4) = 0$   $&hArr;(3x-1)(x+2)=0$   $&rArr;$  ( left[  begin{array}{l}x= dfrac{1}{3}  x=-2 end{array}  right. )      Vậy $x$ $&isin;$ {-2;1/3}.   $e$) $x^2 + 3x - 10=0$   $&hArr; x^2 + 5x - 2x - 10 = 0$   $&hArr; x(x+5) - 2(x+5) = 0$   $&hArr; (x-2)(x+5) = 0$   $&rArr;$  ( left[  begin{array}{l}x=2  x=-5 end{array}  right. )     Vậy $x$ $&isin;$ {-5;2}.

myngocla · Answer

#LC      a,  x^2 - 25 - (x+5) =0   &hArr;(x^2 - 5^2)-(x+5)=0   &hArr; (x+5)(x-5)-(x+5)=0   &hArr; (x+5)(x-5-1)=0   &hArr;(x+5)(x-6)=0   &hArr; ( left[  begin{array}{l}x=-5  x=6 end{array}  right. )     b  (10x+9)x - (5x-1)(2x+3)=0    &hArr;(10x^2+9x)-(10x^2+13x-3)=0     &hArr;10x^2+9x-10x^2 - 13x+3=0     &hArr;-4x + 3 =0     &hArr;x = 3/4     c,  (2x+3)(x-1)+(2x-3)(1-x)=0    &hArr;(2x+3)(x-1)-(2x-3)(x-1)=0     &hArr;(x-1)[2x+3-(2x-3)]=0     &hArr;(x-1).6=0     &hArr;x-1=0 &hArr; x=1     d,  (5x-4)^2 - 49x^2 =0    &hArr; (5x-4)^2-(7x)^2=0     &hArr;(5x-4-7x)(5x-4+7x)=0     &hArr;(-2x-4)(12x-4)=0     &hArr; ( left[  begin{array}{l}-2x-4=0  12x-4=0 end{array}  right. )      &hArr; ( left[  begin{array}{l}x=-2  x= frac{1}{3} end{array}  right. )      e,  x^2+3x-10=0    &hArr;(x^2-2x)+(5x-10)=0     &hArr;x(x-2)+5(x-2)=0     &hArr;(x-2)(x+5)=0     &hArr; ( left[  begin{array}{l}x=2  x=-5 end{array}  right. )