Bài 3: Phương trình bậc hai một ẩn

Từ đề bài ta có:

\(2\sqrt{xy}\le x+y=1\)

\(\Rightarrow xy\le\dfrac{1}{4}\)

Ta có:

\(P=\left(1-\dfrac{1}{x^2}\right)\left(1-\dfrac{1}{y^2}\right)=\dfrac{1-x^2-y^2+x^2y^2}{x^2y^2}\)

\(=1+\dfrac{-\left(x+y\right)^2+2xy+1}{x^2y^2}\)

\(=1+\dfrac{2}{xy}\ge1+8=9\)

Vậy GTNN là A = 9 khi \(x=y=\dfrac{1}{2}\)

Từ \(\dfrac{1}{a}+\dfrac{1}{b}=2\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=2\Rightarrow\dfrac{a+b}{ab}=2\)

\(\Rightarrow2ab=a+b\ge2\sqrt{ab}\)\(\Rightarrow\left\{{}\begin{matrix}ab\ge1\\a+b\ge2\sqrt{ab}\ge2\end{matrix}\right.\)

Áp dụng BĐT AM-GM ta có:

\(a^4+b^2+2ab^2\ge2\sqrt{a^4b^2}+2ab^2=2a^2b+2ab^2\)

\(b^4+a^2+2a^2b\ge2\sqrt{a^2b^4}+2a^2b=2ab^2+2a^2b\)

\(\Rightarrow Q\le\dfrac{1}{2a^2b+2ab^2}+\dfrac{1}{2ab^2+2a^2b}\le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

Đẳng thức xảy ra khi \(a=b=1\)

\(x^2-mx+m+1=0\left(1\right)\)

Xét phương trình (1), áp dụng hệ thức Vi-ét ta có:

\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=m+1\end{matrix}\right.\)

Theo đề bài ta có:
\(x_1^2+x_2^2=6\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=6\)

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

\(\Leftrightarrow m^2-2m-8=0\)

\(\Leftrightarrow\left(m-4\right)\left(m+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m-4=0\\m+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=4\\m=-2\end{matrix}\right.\)

Vậy để phương trình đã cho có 2 nghiệm thỏa mãn \(x_1^2+x_2^2=6\) thì m=4 hoặc m=-2

x² - mx + m +1 =0

xét pt trên áp dụng hệ thức vi-ét ta có:

\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=m+1\end{matrix}\right.\)

Theo đè ta có x_1²+x_1²=6

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=6\)

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

\(\Leftrightarrow m^2-2m-8=0\)

\(\Leftrightarrow\left(m-4\right)\left(m+2\right)=0\)

\(\left[{}\begin{matrix}m-4=0\\m+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=4\\m=-2\end{matrix}\right.\)

vậy pt thỏa mảng x₁²+x₂²=6

\(A=\dfrac{8a^2+b}{4a}+b^2=2a+\dfrac{b}{4a}+b^2\)

\(\ge2a+\dfrac{1-a}{4a}+b^2=2a+\dfrac{1}{4a}-\dfrac{1}{4}+b^2\)

\(\ge a+\dfrac{1}{4a}-b+b^2+\dfrac{3}{4}\)

\(=\left(a+\dfrac{1}{4a}\right)+\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{2}\)

\(=\left(a+\dfrac{1}{4a}\right)+\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)

\(\ge1+\dfrac{1}{2}=\dfrac{3}{2}\)

Dấu = xảy ra khi \(a=b=\dfrac{1}{2}\)

Tìm giá trị nhỏ nhất của : P=$\frac{8a^{2}+b}{4a}+b^{2}$ - Bất đẳng thức và cực trị - Diễn đàn Toán học

Thấy x = 0 không phải là n₀ của pt

=> pt <=> \(\dfrac{4}{x-8+\dfrac{7}{x}}\) +\(\dfrac{5}{x-10+\dfrac{7}{x}}\) = -1

Đặt x - 9 + \(\dfrac{7}{x}\) = a

=> pt <=> \(\dfrac{4}{a+1}\) + \(\dfrac{5}{a-1}\) = -1

<=> \(\dfrac{9a+1}{\left(a-1\right)\left(a+1\right)}\) = -1

<=> \(\dfrac{9a+1}{a^2-1}\) = -1

<=> 9a + 1 = 1 - a²

<=> a² + 9a = 0

<=> a(a + 9) = 0

TH1 a = 0 => x - 9 + \(\dfrac{7}{x}\) = 0

<=> x² - 9x + 7 = 0

<=> ( x - \(\dfrac{9}{2}\) )² = \(\dfrac{53}{4}\)

<=> x = \(\dfrac{9\pm\sqrt{53}}{2}\)

TH2 a = -9 => x - 9 + \(\dfrac{7}{x}\) = -9

<=> x² - 9x + 7 = -9x

<=> x² + 7 = 0 (vô lý)

Vậy x = \(\dfrac{9\pm\sqrt{53}}{2}\)

Ta có:

\(\sqrt{4x^2-4x+1}\)=\(\sqrt{\left(2x-1\right)^2}\)=\(\left|2x-1\right|\)=2015 \(\Rightarrow\) \(\left[{}\begin{matrix}2x-1=2015\\2x-1=-2015\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=2016\\2x=-2014\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1008\\x=-1007\end{matrix}\right.\)

Vậy S=\(\left\{-1007;1008\right\}\)

ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)

\(\Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{xz+yz+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{\left(y+z\right)\left(x+z\right)}{xyz\left(x+y+z\right)}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\x+z=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^8=\left(-y\right)^8\\y^9=\left(-z\right)^9\\z^{10}=\left(-x\right)^{10}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^8-y^8=0\\y^9+z^9=0\\x^{10}-z^{10}=0\end{matrix}\right.\)\(\Rightarrow\left(x^8-y^8\right)\left(y^9+z^9\right)\left(z^{10}-x^{10}\right)=0\)

\(\Rightarrow M=\dfrac{3}{4}\)

pt có 2 no trái dấu khi ac<0

=> m-1>0=>m>1

để phương trình có nghiệm thì: \(\Delta'\ge0\Leftrightarrow3+m\ge0\Leftrightarrow m\ge-3\)

để ptrình có nghiệm trái dấu thì \(m-1>0\Rightarrow m>1\)