\(x^2-2\left(m-1\right)x+m^2+4=0\)

\(\Delta=b^2-4ac\)

\(\Delta=-8m-12\)

Để phương trình có 2 nghiệm phân biệt

\(\Rightarrow\Delta>0\Leftrightarrow m< -\dfrac{3}{2}\)

Theo định lý Viet

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x_1+x_2\right)^2=4\left(m-1\right)^2\\x_1x_2=m^2+4\end{matrix}\right.\)

Theo yêu cầu đề bài \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=3\)

\(\Leftrightarrow\dfrac{x^2_1+x^2_2}{x_1x_2}=3\)

\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=3\)

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

\(\Leftrightarrow\dfrac{4\left(m^2-2m+1\right)-2m^2-8}{m^2+4}=3\)

\(\Leftrightarrow\dfrac{2m^2-8m-4}{m^2+4}=3\)

\(\Leftrightarrow2m^2-8m-4=3m^2+12\)

\(\Leftrightarrow m^2+8m+16=0\)

\(\Delta=b^2-4ac\)

\(\Delta=0\)

\(\Rightarrow m=-\dfrac{b}{2a}=-4\)

a )

\(\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)\)

\(\Leftrightarrow\dfrac{15x-\left(x^2+3x-4\right)}{x^2+3x-4}=\dfrac{12}{x+4}+\dfrac{12}{3x-3}\)

\(\Leftrightarrow\dfrac{12x-x^2+4}{x^2+4x-x-4}=\dfrac{48x+12}{\left(x+4\right)\left(3x-3\right)}\)

\(\Leftrightarrow\dfrac{12x-x^2+4}{x\left(x+4\right)-\left(x+4\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{12x-x^2+4}{\left(x+4\right)\left(x-1\right)}=\dfrac{48x+12}{3\left(x+4\right)\left(x-1\right)}\)

\(\Leftrightarrow12x-x^2+4=\dfrac{48x+12}{3}\)

\(\Leftrightarrow12x-x^2+4=16x+4\)

\(\Leftrightarrow x^2+8x=0\)

\(\Delta=b^2-4ac\)

\(\Delta=64\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-8+\sqrt{64}}{2}=0\left(nhận\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-8-\sqrt{64}}{2}=-8\left(loại\right)\end{matrix}\right.\)

Do \(x=-8\) không thỏa mãn phương trình

Vậy \(x=0\)

b)

\(\dfrac{1}{x-1}+\dfrac{1}{x-2}=\dfrac{1}{x+2}+\dfrac{1}{x+1}\)

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

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

\(\Leftrightarrow\dfrac{2}{x^2-1}=\dfrac{-4}{x^2-4}\)

\(\Leftrightarrow2x^2-8=-4x^2+4\) ( điều kiện \(x\ne\pm1,x\ne\pm2\) )

\(\Leftrightarrow6x^2=12\)

\(\Rightarrow x=\pm\sqrt{2}\)

Bài 3

a) \(x^2-5x+m-2=0\)

Thay \(m=-4\) vào phương trình

\(\Rightarrow x^2-5x-6=0\)

\(\Delta=b^2-4ac\)

\(\Delta=49\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+\sqrt{49}}{2}=6\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-\sqrt{49}}{2}=-1\end{matrix}\right.\)

b )

\(x^2-5x+m-2=0\)

\(\Delta=b^2-4ac\)

\(\Delta=33-4m\)

Theo định lý Viet

\(\Rightarrow\left\{{}\begin{matrix}P=x_1+x_2=\dfrac{-b}{a}\\S=x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}P=x_1+x_2=5\\S=x_1x_2=m-2\end{matrix}\right.\)

Để phương trình có 2 nghiệm dương phân biệt

\(\Rightarrow\left\{{}\begin{matrix}\Delta>0\\P>0\\S>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}33-4m>0\\m-2>0\\5>0\left(đúng\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m< \dfrac{33}{4}\\m>2\end{matrix}\right.\)

\(\Rightarrow2< m< \dfrac{33}{4}\)

Ta có \(2\left(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}\right)=3\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{x_1x_2}}=\dfrac{3}{2}\)

\(\Leftrightarrow\left(\dfrac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{x_1x_2}}\right)^2=\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x_1}+\sqrt{x_2}\right)^2}{x_1x_2}=\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{5+\sqrt{m-2}}{m-2}=\dfrac{9}{4}\)

\(\Leftrightarrow20+4\sqrt{m-2}=9m-18\)

\(\Leftrightarrow4\sqrt{m-2}=9m-38\)

\(\Leftrightarrow64m-128=\left(9m-38\right)^2\)

\(\Leftrightarrow64m-128=81m^2-684m+1444\)

\(\Leftrightarrow81m^2-748m+1572=0\)

\(\Delta=b^2-4ac\)

\(\Delta=50176\)

\(\Rightarrow\left\{{}\begin{matrix}m_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{748+\sqrt{50176}}{162}=6\\m_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{748-\sqrt{50176}}{162}=\dfrac{262}{81}\end{matrix}\right.\)

Vì \(2< m< \dfrac{33}{4}\)

\(\Rightarrow m\in\left\{6;\dfrac{262}{81}\right\}\)

Bài 2

a) \(x^4-24x^2-25=0\) ( 1 )

Đặt \(t=x^2\) ( điều kiện \(t\ge0\) )

\(pt\left(1\right)\Leftrightarrow t^2-24t-25=0\)

\(\Delta=b^2-4ac\)

\(\Delta=676\)

\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{24+\sqrt{676}}{2}=25\left(nhận\right)\\t_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{24-\sqrt{676}}{2}=-1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2=25\)

\(\Rightarrow x=\pm5\)

b)

\(\left\{{}\begin{matrix}2x-y=2\\9x+8y=34\end{matrix}\right.\)

Xét \(2x-y=2\)

\(\Rightarrow x=\dfrac{2+y}{2}\)

Ta có \(9x+8y=34\)

\(\Leftrightarrow\dfrac{9\left(2+y\right)}{2}+8y=34\)

\(\Leftrightarrow\dfrac{18+9y}{2}+8y=34\)

\(\Leftrightarrow\dfrac{18+25y}{2}=34\)

\(\Leftrightarrow18+25y=68\)

\(\Rightarrow y=2\)

\(\Rightarrow x=\dfrac{y+2}{2}=2\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Ta có \(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\Rightarrow\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{3b}=\dfrac{1}{3}\\\dfrac{b}{3c}=\dfrac{1}{3}\\\dfrac{c}{3d}=\dfrac{1}{3}\\\dfrac{d}{3a}=\dfrac{1}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3a=3b\\3b=3c\\3c=3d\\3d=3a\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=d\) ( đpcm )

\(x\in\)N*

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{9}{2}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}=\dfrac{9}{2}\)

Vậy \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{9}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)

1 )

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{{}\begin{matrix}a+1\ge2\sqrt{a}\\b+1\ge2\sqrt{b}\\a+c\ge2\sqrt{ac}\\b+c\ge2\sqrt{bc}\end{matrix}\right.\)

\(\Rightarrow VT\ge2\sqrt{a}.2\sqrt{b}.2\sqrt{ac}.2\sqrt{bc}=16abc\)

\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(a+c\right)\left(b+c\right)\ge16abc\)( đpcm )

\(\dfrac{a^2}{b+2}+\dfrac{b^2}{c+2}+\dfrac{c^2}{a+2}\ge1\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow\dfrac{a^2}{b+2}+\dfrac{b^2}{c+2}+\dfrac{c^2}{a+2}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{9}{9}=1\)

Vậy \(\dfrac{a^2}{b+2}+\dfrac{b^2}{c+2}+\dfrac{c^2}{a+2}\ge1\) ( đpcm )