Toán Lớp 8: 1 . Tìm gtnn và gtln ( nếu được ) a ) A=2x(x-3)-5 b). B=4-3x-2x^2 2. Tìm x biết a. 5x(x-2)-2x+4+0 B.x(x+3)=x^2+6x+9 c.3x(x-2)-4(2-x)=

Question

Toán Lớp 8:  1 . Tìm gtnn và gtln ( nếu được ) a ) A=2x(x-3)-5 b). B=4-3x-2x^2 2. Tìm x biết  a. 5x(x-2)-2x+4+0 B.x(x+3)=x^2+6x+9  c.3x(x-2)-4(2-x)=0 d.5x^2-5x-3(1-x)=0 ( không làm tắt bài tìm x nha mn)

trucoanh55 · Answer

1.   a)A=2x(x-3)-5   =2x&sup2;-6x-5   =2(x&sup2;-3x-5/2)   =2(x&sup2;-3x+9/4-19/4)   =2(x&sup2;-3x+9/4)-19/2   =2[x&sup2;-2.x. 3/2+(3/2)^2]-19/2   =2(x-3/2)^2-19/2   Ta c&oacute;:(x-3/2)^2&ge;0&forall;x   &rArr;2(x-3/2)^2&ge;0&forall;x   &rArr;2(x-3/2)^2-19/2&ge;-19/2&forall;x   Vậy A_(min)=-19/2 khi x-3/2=0&hArr;x=3/2   b)B=4-3x-2x&sup2;   =-2(x&sup2;+3/2x-2)   =-2(x&sup2;+3/2x+9/16-41/16)   =-2(x&sup2;+3/2x+9/16)+41/8   =-2[x&sup2;+2.x. 3/4+(3/4)^2]+41/8   =-2(x+3/4)^2+41/8   Ta c&oacute;:(x+3/4)^2&ge;0&forall;x   &rArr;2(x+3/4)^2&ge;0&forall;x   &rArr;-2(x+3/4)^2&le;0&forall;x   &rArr;-2(x+3/4)^2+41/8&le;41/8&forall;x   Vậy B_(max)=41/8 khi x+3/4=0&hArr;x=-3/4   2.   a)5x(x-2)-2x+4=0   &hArr;5x(x-2)-(2x-4)=0   &hArr;5x(x-2)-2(x-2)=0   &hArr;(x-2)(5x-2)=0   (1)x-2=0&hArr;x=2   (2)5x-2=0&hArr;x=2/5   Vậy x&isin;{2;2/5}   b)x(x+3)=x&sup2;+6x+9   &hArr;x(x+3)=x&sup2;+2.x.3+3&sup2;   &hArr;x(x+3)=(x+3)&sup2;   &hArr;x(x+3)-(x+3)&sup2;=0   &hArr;(x+3)[x-(x+3)]=0   &hArr;(x+3)(x-x-3)=0   &hArr;(x+3).(-3)=0   &hArr;x+3=0   &hArr;x=-3   Vậy x=-3   c)3x(x-2)-4(2-x)=0   &hArr;3x(x-2)+4(x-2)=0   &hArr;(x-2)(3x+4)=0   (1)x-2=0&hArr;x=2   (2)3x+4=0&hArr;x=-4/3   Vậy x&isin;{2;-4/3}   d)5x&sup2;-5x-3(1-x)=0   &hArr;(5x&sup2;-5x)+3(x-1)=0   &hArr;5x(x-1)+3(x-1)=0   &hArr;(x-1)(5x+3)=0   (1)x-1=0&hArr;x=1   (2)5x+3=0&hArr;x=-3/5   Vậy x&isin;{1;-3/5}

lanngocha · Answer

Giải đáp+Lời giải và giải thích chi tiết:    1)   a) A = 2x(x - 3) - 5            = 2x&sup2; - 6x - 5            = 2(x&sup2; - 3x +  frac{9}{4}) -  frac{19}{2}            = 2(x -  frac{3}{2})&sup2; -  frac{19}{2}   V&igrave;  2(x -  frac{3}{2})&sup2; &ge; 0 với &forall;x    n&ecirc;n  2(x -  frac{3}{2})&sup2; -  frac{19}{2} &ge; -  frac{19}{2}   Dấu '=' xảy ra khi x =  frac{3}{2}   Vậy min A =  frac{-19}{2} khi x =  frac{3}{2}   b) B = 4 - 3x - 2x&sup2;          = - 2(x&sup2; +  frac{3}{2} +  frac{9}{16}) +  frac{25}{4}          = -2(x +  frac{3}{4})&sup2; +  frac{25}{4}   V&igrave; -2(x +  frac{3}{4})&sup2; &le; 0 với &forall;x   n&ecirc;n -2(x +  frac{3}{4})&sup2; +  frac{25}{4} &le;  frac{25}{4}   Dấu '=' xảy ra khi x =  frac{-3}{4}   Vậy max B =  frac{25}{4} khi x =  frac{-3}{4}   2)   a) 5x(x - 2) - 2x + 4 = 0   &hArr; 5x(x - 2) - (2x - 4) = 0   &hArr; 5x(x - 2) - 2(x - 2) = 0   &hArr; (x - 2)(5x - 2) = 0   &rArr; x = 2   hoặc x =  frac{2}{5}   b) x(x + 3) = x&sup2; + 6x + 9   &hArr; x(x + 3) = (x + 3)&sup2;   &hArr; x(x + 3) - (x + 3)&sup2; = 0   &hArr; (x + 3)(x - x - 3) = 0   &hArr; -3(x + 3) = 0   &hArr; x + 3 = 0   &hArr; x = - 3   c) 3x(x - 2) - 4(2 - x) = 0   &hArr; 3x(x - 2) + 4(x - 2) = 0   &hArr; (x - 2)(3x + 4) = 0   &rArr; x = 2   hoặc  x =  frac{-4}{3}   d) 5x&sup2; - 5x - 3(1 - x) = 0   &hArr; 5x(x - 1) - 3(1 - x) = 0   &hArr; 5x(x - 1) + 3(x - 1) = 0   &hArr; (x - 1)(5x + 3) = 0   &rArr; x = 1   hoặc x =  frac{-3}{5}