a: \(A=-\left(x^2-4x-3\right)\)

\(=-\left(x^2-4x+4-7\right)\)

\(=-\left(x-2\right)^2+7< =7\)

Dấu '=' xảy ra khi x=2

b: \(B=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)

Dấu '=' xảy ra khi x=1/2

c: \(C=-2\left(x^2-x+\dfrac{5}{2}\right)\)

\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)

Dấu '=' xảy ra khi x=1/2

e: \(E=-\left(x^2+6x+9+1\right)=-\left(x+3\right)^2-1< =-1\)

Dấu = xảy ra khi x=-3

a) A = \(2x^2+x-1=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)\)\(-\frac{9}{8}=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\)

Vì \(\left(x+\frac{1}{4}\right)^2\ge0\forall x\Leftrightarrow2\left(x+\frac{1}{4}\right)^2\ge0\forall x\Leftrightarrow2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\forall x\Leftrightarrow A\ge-\frac{9}{8}\)

Dấu = xảy ra \(\Leftrightarrow\)\(x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)

Vậy minA =\(-\frac{9}{8}\)khi \(x=-\frac{1}{4}\).

b) B=\(5x-3x^2+2=-3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)+\frac{49}{12}=-3\left(x-\frac{5}{6}\right)^2+\frac{49}{12}\)

Vì \(\left(x-\frac{5}{6}\right)^2\ge0\forall x\Leftrightarrow-3\left(x-\frac{5}{6}\right)^2\le0\forall x\Leftrightarrow-3\left(x-\frac{5}{6}\right)^2+\frac{49}{12}\le\frac{49}{12}\forall x\Leftrightarrow B\le\frac{49}{12}\forall x\)

Dấu = xảy ra \(\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

Vậy maxB = \(\frac{49}{12}\)khi \(x=\frac{5}{6}\).

\(\frac{x^2+15}{x^2+3}\)

\(=\frac{x^2+3+12}{x^2+3}\)

\(=\frac{x^2+3}{x^2+3}+\frac{12}{x^2+3}\)

\(=1+\frac{12}{x^2+3}\)

\(x^2\ge0\)

\(x^2+3\ge3\)

\(\frac{12}{x^2+3}\le4\)

\(1+\frac{12}{x^2+3}\le5\)

ĐS: 5

a) \(x^2+7x+12\)

\(=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

b) \(3x^2-5x+2\)

\(=3x^2-3x-2x+2\)

\(=3x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(3x-2\right)\)

a) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 4)(x + 3)

b) 3x² - 5x + 2 = 3x² - 3x - 2x + 2 = 3x(x - 1) - 2(x - 1) = (3x - 2)(x - 1)

c) x² + 9x - 10 = x² + 10x - x - 10 = x(x + 10) - (x + 10) = (x - 1)(x + 10)

d) x² - 7x - 8 = x² - 8x + x - 8 = x(x - 8) + (x - 8) = (x + 1)(x - 8)

e) 2x² + 3x - 5 = 2x² + 5x - 2x - 5 = x(2x + 5) - (2x + 5) = (x  - 1)(2x + 5)

a: \(\Leftrightarrow\left(3x+2\right)\left(5-x\right)=-9x^2+4\)

\(\Leftrightarrow\left(3x+2\right)\left(5-x\right)+\left(3x+2\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(2x+3\right)=0\)

=>x=-2/3 hoặc x=-3/2

b: \(\Leftrightarrow4x\left(x+5\right)+x^2-25=0\)

\(\Leftrightarrow\left(x+5\right)\left(5x-5\right)=0\)

=>x=-5 hoặc x=1

c: \(\Leftrightarrow3x\left(x-1\right)=\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

=>x=1 hoặc x=-1/2

D

help

D

ai đó dịch hộ mình @@

tìm giá trị của n sao cho đa thức 2x ^ 5 - 3x ^ 3 + x ^ 2 + n chia hết cho đa thức x + 2