Bài 7: Phương trình quy về phương trình bậc hai

Ace LegonaHoàng Ngọc Anh

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

đặt x -3 =t => x =t+3

f(t) =2(t+3) +1/t^2 +4

f(t) =2t +1/t^2 +10

f(t) =(1/t^2 -2 +t^2 ) -(t^2 -2t +1) +13

f(t) =(1/t-t)^2 -(t-1)^2 +13

f(t) >= 13

đẳng thức khi t=1 => x =4

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

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

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

Vì \(A^2+A+6>0\)

\(\Rightarrow A=1\)

\(x^2+\sqrt{x+2}=2\left(x\ge-2\right)\)

\(\Leftrightarrow\left(x^2-1\right)+\left(\sqrt{x+2}-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+\dfrac{x+2-1}{\sqrt{x+2}-1}=0\)

\(\Leftrightarrow\left[\left(x-1\right)+\dfrac{1}{\sqrt{x+2}-1}\right]\left(x+1\right)=0\)

Pt \(\left(x-1\right)+\dfrac{1}{\sqrt{x+2}-1}=0\) vô n_o

=> x + 1 = 0

<=> x = - 1 (nhận)

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

Đk:\(x\ge -2\)

\(pt\Leftrightarrow\sqrt{x+2}=2-x^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\x+2=x^4-4x^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\-x^4+4x^2+x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\-\left(x-2\right)\left(x+1\right)\left(x^2+x-1\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\sqrt{2}\le x\le\sqrt{2}\\\left[{}\begin{matrix}x=2\left(loai\right)\\x=-1\\x=\dfrac{\sqrt{5}-1}{2}\\x=-\dfrac{\sqrt{5}+1}{2}\left(loai\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Delta\)' = 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 = \(-\dfrac{1}{2}\)

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

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

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

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

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 \(A_{Min}=\dfrac{7}{16}\)

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

2)a)\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)

c)\(a^3+b^3-a^2b-ab^2=a^2\left(a-b\right)-b^2\left(a-b\right)=\left(a-b\right)^2\left(a+b\right)\ge0\\ \Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)

b)\(a^3+b^3\ge a^2b+ab^2\Leftrightarrow4a^3+4b^3\ge a^3+b^3+3a^b+3ab^2\\ \Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\Leftrightarrow\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{2}\right)^3\)

\(x+2y=4\Leftrightarrow x=4-2y\)

\(\Rightarrow xy=y\left(4-2y\right)=-2y^2+4y=-2\left(y-1\right)^2+2\le2\)

Vậy max M là 2 khi y=1, x= 2

2)Tương tự