\(a.A=\dfrac{3-x}{x^2-2}\)

\(\Leftrightarrow Ax^2-2A-3+x=0\)

\(\Leftrightarrow Ax^2+x-2A-3=0\)

\(\Delta=b^2-4ac\ge0\)

\(\Leftrightarrow1-4.A\left(-2A-3\right)\ge0\)

\(\Leftrightarrow1+8A^2+12A\ge0\)

\(\Leftrightarrow8A^2+12A+1\ge0\)

\(\Leftrightarrow\dfrac{-3-\sqrt{7}}{4}\le A\le\dfrac{-3+\sqrt{7}}{4}\)

Suy ra: \(Min_A=\dfrac{-3-\sqrt{7}}{4}\Leftrightarrow x=\dfrac{-b}{2A}=\dfrac{-1}{2.\dfrac{-3-\sqrt{7}}{4}}=3-\sqrt{7}\)

\(Max_A=\dfrac{-3+\sqrt{7}}{4}\Leftrightarrow x=\dfrac{-b}{2A}=\dfrac{-1}{2.\dfrac{-3+\sqrt{7}}{4}}=3+\sqrt{7}\)

\(b.B=\dfrac{x^2-x}{x^2+1}\)

\(\Leftrightarrow Bx^2+B-x^2+x=0\)

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

\(\Delta=b^2-4ac=1^2-4.B.\left(B-1\right)\)

\(=1-4B^2+4B\)

\(\Leftrightarrow\dfrac{1-\sqrt{2}}{2}\le B\le\dfrac{1+\sqrt{2}}{2}\)

\(\Leftrightarrow Min_B=\dfrac{1-\sqrt{2}}{2}\Leftrightarrow x=\dfrac{-b}{2B}=\dfrac{-1}{2.\dfrac{1-\sqrt{2}}{2}}=1+\sqrt{2}\)

\(Max_B=\dfrac{1+\sqrt{2}}{2}\Leftrightarrow x=\dfrac{-b}{2B}=\dfrac{-1}{2.\dfrac{1+\sqrt{2}}{2}}=1-\sqrt{2}\)

P/S: Mk làm thế nhưng khi thử thay x vào thì không đúng, bn xem lại giúp nha

Biểu thức nào em?

a) \(A=x^2-2x+5\)

\(A=x^2-2x+1+4\)

\(A=\left(x-1\right)^2+4\)

Có:  \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\)

Dấu '=' xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)

Vậy: \(Min_A=4\) tại \(x=1\)

b) \(B=x^2+x+1\)

\(B=x^2+x+\frac{1}{4}+\frac{3}{4}\)

\(B=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Có: \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu '=' xảy ra khi: \(\left(x+\frac{1}{2}\right)^2=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

Vậy: \(Min_B=\frac{3}{4}\) tại \(x=-\frac{1}{2}\)

c) \(C=4x-x^2+3\)

\(C=-x^2+4x-4+8\) 

\(C=8-\left(x^2-4x+4\right)\)

\(C=8-\left(x-2\right)^2\)

Có: \(\left(x-2\right)^2\ge0\Rightarrow8-\left(x-2\right)^2\le8\)

Dấu '=' xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

Vậy: \(Max_C=8\) tại \(x=2\)

bây h đi học rồi -_-

1: \(A=\left(x-1\right)^2-10\ge-10\)

Dấu '=' xảy ra khi x=1

2: \(B=-\left|x-1\right|-2\cdot\left(2y-1\right)^2+100\le100\)

Dấu '=' xảy ra khi x=1 và y=1/2

`(x-1)^2 >=0 => (x-1)^2 - 10 >= -10`

Dấu bằng xảy ra khi `x = 1`.

Vì `-|x-1| <=0, -2(2y-1)^2 <= 0`

`=> -|x-1| - 2(2y-1)^2 + 100 <= 100`

Dấu bằng xảy ra `<=> x = 1, y = 1/2`.

a, Ta có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu " = " khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(MIN_A=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

b, Để B lớn nhất thì \(\left(x-\dfrac{2}{3}\right)^2+9\) nhỏ nhất

Ta có: \(\left(x-\dfrac{2}{3}\right)^2+9\ge9\)

\(\Leftrightarrow B=\dfrac{4}{\left(x-\dfrac{2}{3}\right)^2+9}\le\dfrac{4}{9}\)

Dấu " = " khi \(\left(x-\dfrac{2}{3}\right)^2=0\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(MAX_B=\dfrac{4}{9}\) khi \(x=\dfrac{2}{3}\)

a. A=\(\dfrac{-2}{x^{2^{ }}-2x+5}\)= \(\dfrac{-2}{\left(x-1\right)^{2^{ }}+4}\)

Ta có: (x-1) ² ≥ 0 với mọi x

⇔ (x- 1)² +4 ≥4

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

Dấu''='' xảy ra ⇔ x-1=0

⇔x=1

Vậy maxA= -0,5 ⇔ x=1

b. B=\(\dfrac{3}{x^{2^{ }}-2x+1}\)=\(\dfrac{3}{\left(x-1\right)^2}\)

Ta có: (x-1)² ≥ 0 với mọi x

⇔ \(\dfrac{3}{\left(x-1\right)^2}\)≤0