mik cần gấp giúp vs ạ

\(C=4x^2+y^2-4x+8y+12\)

\(C=4x^2-4x+1+y^2+8y+16-5\)
\(C=\left(4x^2-4x+1\right)+\left(y^2+8y+16\right)-5\)

\(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\)

Mà: \(\left\{{}\begin{matrix}\left(2x-1\right)^2\ge0\forall x\\\left(y+4\right)^2\ge0\forall x\end{matrix}\right.\)

Nên: \(C=\left(2x-1\right)^2+\left(y+4\right)^2-5\ge-5\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2x-1=0\\y+4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

Vậy: \(C_{min}=-5\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-4\end{matrix}\right.\)

\(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\left(1\right)\)

a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)

\(\left(1\right)\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)

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

b) Để \(A=-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{x^2}{x+1}=-\dfrac{1}{2}\left(x\ne-1\right)\)

\(\Leftrightarrow2x^2=-\left(x+1\right)\)

\(\Leftrightarrow2x^2+x+1=0\)

\(\Delta=1-8=-7< 0\)

Nên phương trình trên vô nghiệm \(\left(x\in\varnothing\right)\)

c) Để \(A< 1\) 

\(\Leftrightarrow\dfrac{x^2}{x+1}< 1\)

\(\Leftrightarrow x^2< x+1\left(x\ne-1\right)\)

\(\Leftrightarrow x^2-x-1< 0\)

\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}< 0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2< \dfrac{5}{4}\)

\(\Leftrightarrow-\dfrac{\sqrt[]{5}}{2}< x-\dfrac{1}{2}< \dfrac{\sqrt[]{5}}{2}\)

\(\Leftrightarrow\dfrac{-\sqrt[]{5}+1}{2}< x< \dfrac{\sqrt[]{5}+1}{2}\)

d) Để A nguyên

\(\Leftrightarrow\dfrac{x^2}{x+1}\in Z\)

\(\Leftrightarrow x^2⋮x+1\)

\(\Leftrightarrow x^2-x\left(x+1\right)⋮x+1\)

\(\Leftrightarrow x^2-x^2+x⋮x+1\)

\(\Leftrightarrow x⋮x+1\)

\(\Leftrightarrow x-x-1⋮x+1\)

\(\Leftrightarrow-1⋮x+1\)

\(\Leftrightarrow x+1\in\left\{-1;1\right\}\)

\(\Leftrightarrow x\in\left\{-2;0\right\}\left(x\in Z\right)\)

!ERROR 404!

a) \(4x^2+16x+3=0\)

\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)

b) \(7x^2+16x+2=1+3x^2\)

\(4x^2+16x+1=0\)

\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)

c) \(4x^2+20x+4=0\)

\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)

\(\Leftrightarrow x^2+5x+1=0\)

\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)

\(G=-3x^2-5x+1\\ =-3\left(x^2+2.\dfrac{5}{6}x+\dfrac{25}{36}\right)+\dfrac{37}{12}\\ =\dfrac{37}{12}-3\left(x+\dfrac{5}{6}\right)^2\\ Vì:\left(x+\dfrac{5}{6}\right)^2\ge0\forall x\in R\\ Vậy:G_{max}=\dfrac{37}{12}.khi.x=-\dfrac{5}{6}\)

Lời giải:

$-F=5x^2+4x-3=5(x^2+\frac{4}{5}x+\frac{2^2}{5^2})-\frac{19}{5}$

$=5(x-\frac{2}{5})^2-\frac{19}{5}\geq \frac{-19}{5}$ với mọi $x$

$\Rightarrow F\leq \frac{19}{5}$

Vậy $F_{\max}=\frac{19}{5}$. Giá trị này đạt tại $x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}$

E = - 3\(x^2\) - \(x\) + 2

E = - 3.( \(x^2\) + 2.\(\dfrac{1}{6}\)\(x\) + \(\dfrac{1}{36}\)) + 2

E = -3.(\(x\) + \(\dfrac{1}{6}\))² + \(\dfrac{25}{12}\)

Vì (\(x+\dfrac{1}{6}\))² ≥ 0 ∀ \(x\) ⇒ -3.(\(x+\dfrac{1}{6}\))² ≤ 0 ⇒ -3(\(x+\dfrac{1}{6}\))² + \(\dfrac{25}{12}\) ≤ \(\dfrac{25}{12}\)

E_max = \(\dfrac{25}{12}\) ⇔ \(x\) = - \(\dfrac{1}{6}\)