I:
a: \(=x^2-2x+1+x^2-4x+4\)

\(=2x^2-6x+5\)

\(=2\left(x^2-3x+\dfrac{5}{2}\right)\)

\(=2\left(x^2-3x+\dfrac{9}{4}+\dfrac{1}{4}\right)\)

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

Dấu = xảy ra khi x=3/2

b: \(=-4\left(x^2-2x+\dfrac{3}{4}\right)\)

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

Dấu = xảy ra khi x=1

mn cố gắng giúp em với

tìm Max thì bn bình phương lên r bunyakovsky

Min thì Áp dụng \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\)

Ta có: A=x²+y²=1-2xy

Vì x+y=1 => x=1-y

Khi đó A=1-2(1-y)y

=1-2y+2y²

=\(2\left(y^2-y+\frac{1}{4}\right)+\frac{1}{2}\)

=\(2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\)

Vif \(2\left(y-\frac{1}{2}\right)^2\ge0\Rightarrow A=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Dấu "=" xảy ra khi y=1/2 <=> x=1/2

Vậy Amin=1/2 khi x=y=1/2

Ta có:

\(A=\dfrac{2010x+2680}{x^2+1}\)

\(=\dfrac{-335x^2-335+335x^2+2010x+3015}{x^2+1}\)

\(=-335+\dfrac{335\left(x+3\right)^2}{x^2+1}\ge-335\)

Vậy \(Min_A=-335\) tại \(x=-3\)

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a) C được xác định <=> x khác +- 2

b) Ta có : \(C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

Để C = 0 thì x - 1 = 0 <=> x = 1 (tm)

c) Để C nhận giá trị dương thì x - 1 > 0 <=> x > 1

Kết hợp với ĐK => Với x > 1 và x khác 2 thì C nhận giá trị dương

Ta có:

\(I=\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|=\left(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{4}\right|\right)+\left|x+\frac{1}{3}\right|\)

\(=\left(\left|x+\frac{1}{2}\right|+\left|-x-\frac{1}{4}\right|\right)+\left|x+\frac{1}{3}\right|\ge\left|x+\frac{1}{2}-x-\frac{1}{4}\right|+\left|x+\frac{1}{3}\right|=\frac{1}{4}+\left|x+\frac{1}{3}\right|\ge\frac{1}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x+\frac{1}{2}\right)\left(-x-\frac{1}{4}\right)\ge0\\x+\frac{1}{3}=0\end{cases}}\Leftrightarrow x=-\frac{1}{3}\)

Vậy min I = 1/4 đạt tại x = -1/3.

Lời giải:

ĐK: $x\neq 2019$

PT $\Rightarrow A(x-2019)^2=2019x$

$\Leftrightarrow Ax^2-x(4038A+2019)+A.2019^2=0(*)$

Vì biểu thức $A$ xác định nên PT $(*)$ có nghiệm.

$\Rightarrow \Delta=(4038A+2019)^2-4A^2.2019^2\geq 0$

$\Leftrightarrow 2019^2(2A+1)^2-4A^2.2019^2\geq 0$

$\Leftrightarrow (2A+1)^2-(2A)^2\geq 0$

$\Leftrightarrow 4A+1\geq 0$

$\Leftrightarrow A\geq -\frac{1}{4}$
Vậy GTNN của $A$ là $\frac{-1}{4}$. $A$ không có GTLN