Bài 7: Phương trình quy về phương trình bậc hai

\(A=x^2+3xy+4y^2\)

\(=\dfrac{9}{16}x^2+3xy+4y^2+\dfrac{7}{16}x^2\)

\(=4\left(\dfrac{9}{64}x^2+\dfrac{3}{4}xy+y^2\right)+\dfrac{7}{16}x^2\)

\(=4\left(\dfrac{3}{8}x+y^2\right)+\dfrac{7}{16}x^2\ge4\cdot0+\dfrac{9}{17}\cdot1^2=\dfrac{7}{16}\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{8}\end{matrix}\right.\)

Vậy với \(\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{8}\end{matrix}\right.\) khi \(A_{Min}=\dfrac{7}{16}\)

Chắc bạn ghi nhầm đề hay sao ấy. 2014 hay 2012 vậy b

Giả sử đề bạn là 2012 thì mình làm nhé.

\(x^4+\sqrt{x^2+2012}=2012\)

\(\Leftrightarrow\left(x^4+x^2+\dfrac{1}{4}\right)=\left(x^2+2012-\sqrt{x^2+2012}+\dfrac{1}{4}\right)\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{2}\right)^2=\left(\sqrt{x^2+2012}-\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow x^2+\dfrac{1}{2}=\sqrt{x^2+2012}-\dfrac{1}{2}\)

\(\Leftrightarrow\left(x^2+2012-\sqrt{x^2+2012}+\dfrac{1}{4}\right)=2011,25\)

\(\Leftrightarrow\left(\sqrt{x^2+2012}-\dfrac{1}{2}\right)^2=2011,25\)

Tới đây thì đơn giản rồi. b làm tiếp nhé

Theo hệ thức Vi-ét a,b là nghiệm của phương trình:

x²-7x+12=0

\(\Delta=\left(-7\right)^2-4.1.12=1>0\)\(\Rightarrow\)phương trình có 2 nghiệm phân biệt

\(x_1=\dfrac{-\left(-7\right)+\sqrt{1}}{2}=4\)

\(x_2=\dfrac{-\left(-7\right)-\sqrt{1}}{2}=3\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\\\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}a+b=7\\ab=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7-b\\\left(7-b\right)=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=7-b\\b^2-7b+12=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7-b\\\left(b-4\right)\left(b-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=3\\a=4\end{matrix}\right.\\\left[{}\begin{matrix}b=4\\b=3\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(a;b\right)=\left(3;4\right)\) hoặc \(\left(a;b\right)=\left(4;3\right)\)

a = 3; b = 4 hoặc a = 4; b = 3

\(A=\sqrt[3]{3+\sqrt{\dfrac{368}{27}}}+\sqrt[3]{3-\sqrt{\dfrac{368}{27}}}\)

\(\Leftrightarrow A^3=6+3A.\sqrt[3]{-\dfrac{125}{27}}=6-5A\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+6\right)=0\)

Vì \(A^2+A+6>0\)

\(\Rightarrow A=1\)

\(P=\dfrac{bc}{a\left(b+c\right)}+\dfrac{ca}{b\left(c+a\right)}+\dfrac{ab}{c\left(a+b\right)}\)

\(=\dfrac{b^2c^2}{abc\left(b+c\right)}+\dfrac{c^2a^2}{abc\left(c+a\right)}+\dfrac{a^2b^2}{abc\left(a+b\right)}\)

\(\ge\dfrac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\ge\dfrac{3abc\left(a+b+c\right)}{2abc\left(a+b+c\right)}=\dfrac{3}{2}\)

Dấu = xảy ra khi \(a=b=c\)

\(\Delta\)' = m² - m² + m - 1 = m - 1

ta có phương trình có nghiệm x₁ ; x₂ \(\Leftrightarrow\) \(\Delta\)' \(\ge\) 0

\(\Leftrightarrow\) m - 1 \(\ge\) 0 \(\Leftrightarrow\) m \(\ge\) 1

ta có : A = x₁² + x₂² = (x₁ + x₂)² - 2x₁x₂

áp dụng hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-m+1\end{matrix}\right.\)

thay : (2m)² - 2(m² - m + 1) = 4m² - 2m² + 2m - 2

= 2m² + 2m - 2 = 2 (m² + m - 1) = 2 (m² + 2.m.\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{4}\) - 1) = 2 [(m + \(\dfrac{1}{2}\))² - \(\dfrac{5}{4}\) ] = 2(m + \(\dfrac{1}{2}\))² - \(\dfrac{5}{2}\) \(\ge\) \(-\dfrac{5}{2}\)

minA = \(\dfrac{-5}{2}\) khi m + \(\dfrac{1}{2}\) = 0 \(\Leftrightarrow\) m = \(-\dfrac{1}{2}\)