Question

Đã trả lời bởi chuyên gia

Answer 1

$a, T h a y x = - 3 v à o B t a c ó : B = \frac{- 3 + 2}{3 (- 3) + 2} = \frac{- 1}{- 7} = \frac{1}{7} V ậ y B = \frac{1}{7} t ạ i x = - 3 b, M = A B = (\frac{1}{x + 2} - \frac{2 x}{4 - x^{2}} + \frac{3}{x - 2}) . \frac{x + 2}{3 x + 2}, đ k : x \neq \pm 2; x \neq - \frac{2}{3} = \frac{x - 2 + 2 x + 3 (x + 2)}{(x - 2) (x + 2)} . \frac{x + 2}{3 x + 2} = \frac{x - 2 + 2 x + 3 x + 6}{x - 2} . \frac{1}{3 x + 2} = \frac{6 x + 4}{x - 2} . \frac{1}{3 x + 2} = \frac{2 (3 x + 2)}{x - 2} . \frac{1}{3 x + 2} = \frac{2}{x - 2} V ậ y M = \frac{2}{x - 2}$

$c, N = M . (x^{3} - x^{2} - 2 x) = \frac{2}{x - 2} . [x (x^{2} - x - 2)] = \frac{2}{x - 2} . x (x - 2) (x + 1) = \frac{2 x (x + 1)}{x - 2} = \frac{2 x^{2} + 2 x}{x - 2} = \frac{2 x^{2} - 4 x + 6 x - 12 + 12}{x - 2} = 2 x + 6 + \frac{12}{x - 2} = 2 (x - 2) + \frac{12}{x - 2} + 10 T a c ó : 2 (x - 2) + \frac{12}{x - 2} \geq 2 \sqrt{2 (x - 2) . \frac{12}{x - 2}} = 4 \sqrt{6}, t h e o C o s i = > N = 2 (x - 2) + \frac{12}{x - 2} + 10 \geq 4 \sqrt{6} + 10 v ớ i m ọ i x D ấ u = x ả y r a < = > x - 2 = \frac{12}{x - 2} < = > {(x - 2)}^{2} = 12 < = > x - 2 = \pm 2 \sqrt{3} < = > x = 2 \pm 2 \sqrt{3} V ậ y x = 2 \pm 2 \sqrt{3} t h ì N c ó G T N N = 4 \sqrt{6} + 10$ ..

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Answer 2

$a, T h a y x = - 3 v à o B t a c ó : B = \frac{- 3 + 2}{3 (- 3) + 2} = \frac{- 1}{- 7} = \frac{1}{7} V ậ y B = \frac{1}{7} t ạ i x = - 3 b, M = A B = (\frac{1}{x + 2} - \frac{2 x}{4 - x^{2}} + \frac{3}{x - 2}) . \frac{x + 2}{3 x + 2}, đ k : x \neq \pm 2; x \neq - \frac{2}{3} = \frac{x - 2 + 2 x + 3 (x + 2)}{(x - 2) (x + 2)} . \frac{x + 2}{3 x + 2} = \frac{x - 2 + 2 x + 3 x + 6}{x - 2} . \frac{1}{3 x + 2} = \frac{6 x + 4}{x - 2} . \frac{1}{3 x + 2} = \frac{2 (3 x + 2)}{x - 2} . \frac{1}{3 x + 2} = \frac{2}{x - 2} V ậ y M = \frac{2}{x - 2}$

$c, N = M . (x^{3} - x^{2} - 2 x) = \frac{2}{x - 2} . [x (x^{2} - x - 2)] = \frac{2}{x - 2} . x (x - 2) (x + 1) = \frac{2 x (x + 1)}{x - 2} = \frac{2 x^{2} + 2 x}{x - 2} = \frac{2 x^{2} - 4 x + 6 x - 12 + 12}{x - 2} = 2 x + 6 + \frac{12}{x - 2} = 2 (x - 2) + \frac{12}{x - 2} + 10 T a c ó : 2 (x - 2) + \frac{12}{x - 2} \geq 2 \sqrt{2 (x - 2) . \frac{12}{x - 2}} = 4 \sqrt{6}, t h e o C o s i = > N = 2 (x - 2) + \frac{12}{x - 2} + 10 \geq 4 \sqrt{6} + 10 v ớ i m ọ i x D ấ u = x ả y r a < = > x - 2 = \frac{12}{x - 2} < = > {(x - 2)}^{2} = 12 < = > x - 2 = \pm 2 \sqrt{3} < = > x = 2 \pm 2 \sqrt{3} V ậ y x = 2 \pm 2 \sqrt{3} t h ì N c ó G T N N = 4 \sqrt{6} + 10$ ..

Answer 3

MỌI NGƯỜI TRẢ LỜI CHƯA ?

Answer 4

Hình như câu c sai

Answer 5

a, Thay x=−3 vào B ta có:B=−3+23(−3)+2=−1−7=17Vậy B=17 tại x=−3b, M=AB=(1x+2−2x4−x2+3x−2).x+23x+2, đk: x≠±2; x≠−23=x−2+2x+3(x+2)(x−2)(x+2).x+23x+2=x−2+2x+3x+6x−2.13x+2=6x+4x−2.13x+2=2(3x+2)x−2.13x+2=2x−2Vậy M=2x−2

c, N=M.(x3−x2−2x)=2x−2.[x(x2−x−2)]=2x−2.x(x−2)(x+1)=2x(x+1)x−2=2x2+2xx−2=2x2−4x+6x−12+12x−2=2x+6+12x−2=2(x−2)+12x−2+10Ta có: 2(x−2)+12x−2≥2√2(x−2).12x−2=4√6, theo Cosi=>N=2(x−2)+12x−2+10≥4√6+10 với mọi xDấu = xảy ra<=>x−2=12x−2<=>(x−2)2=12<=>x−2=±2√3<=>x=2±2√3Vậy x=2±2√3 thì N có GTNN=4√6+10

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