\(\sqrt{3}-\dfrac{m}{n}>0\Leftrightarrow\sqrt{3}>\dfrac{m}{n}\Leftrightarrow3n^2>m^2\)

Vì \(m,n\ge1\) nên \(3n^2\ge m^2+1\)

Với \(3n^2=m^2+1\Leftrightarrow m^2+1⋮3\Leftrightarrow m^2\) chia 3 dư 2 (vô lí)

\(\Leftrightarrow3n^2\ge m^2+2\)

Lại có \(4m^2>1\Leftrightarrow\left(m+\dfrac{1}{2m}\right)^2=m^2+1+\dfrac{1}{4m^2}< m^2+2\)

\(\Leftrightarrow\left(m+\dfrac{1}{2m}\right)^2< 3n^2\Leftrightarrow m+\dfrac{1}{2m}< n\sqrt{3}\\ \Leftrightarrow n\sqrt{3}-m>\dfrac{1}{2m}\)

\(P=\frac{5a+5b+2c}{\sqrt{12\left(a^2+ab+bc+ca\right)}+\sqrt{12\left(b^2+ab+bc+ca\right)}+\sqrt{c^2+ab+bc+ca}}\)

\(=\frac{5a+5b+2c}{\sqrt{12\left(a+b\right)\left(a+c\right)}+\sqrt{12\left(a+b\right)\left(b+c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(=\frac{5a+5b+2c}{\sqrt{\left(6a+6b\right)\left(2a+2c\right)}+\sqrt{\left(6a+6b\right)\left(2b+2c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(\Rightarrow P\ge\frac{2\left(5a+5b+2c\right)}{6a+6b+2a+2c+6a+6b+2b+2c+a+c+b+c}\)

\(\Rightarrow P\ge\frac{2\left(5a+5b+2c\right)}{3\left(5a+5b+2c\right)}=\frac{2}{3}\)

\(P_{min}=\frac{2}{3}\) khi \(\left\{{}\begin{matrix}a=b=1\\c=5\end{matrix}\right.\)

\(\sqrt{3}>\frac{m}{n}\Rightarrow3>\frac{m^2}{n^2}\Rightarrow3n^2>m^2\Rightarrow3n^2\ge m^2+1\)

với 3n²=m²+1=>m²+1 chia hết cho 3

=>m² chia 3 dư 2(vô lí)

\(\Rightarrow3n^2\ge m^2+2\)

lại có:\(\left(m+\frac{1}{2m}\right)^2=m^2+1+\frac{1}{4m^2}< m^2+2\)

\(\Rightarrow\left(m+\frac{1}{2m}\right)^2< 3n^2\Rightarrow m+\frac{1}{2m}< \sqrt{3}n\)

\(\Rightarrow\frac{m}{n}+\frac{1}{2mn}< \sqrt{3}\left(Q.E.D\right)\)

ĐKXĐ:...

\(\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x-2018}-\sqrt{y-2018}+x^2-y^2=0\)

\(\Leftrightarrow\frac{\left(x-y\right)\left(x+y\right)}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\frac{x-y}{\sqrt{x-2018}+\sqrt{y-2018}}+\left(x-y\right)\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\frac{x+y}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\frac{1}{\sqrt{x-2018}+\sqrt{y-2018}}+x+y\right)=0\)

\(\Leftrightarrow x=y\) (ngoặc phía sau luôn dương)

Thay vào M chẳng được cái gì cả, \(M=x^{11}-x^{2018}\) :(

Chắc bạn nhầm đề

trùi s ghim lên đay cx k ai giải v trùi

Δ=(2m-2)^2-4(m-3)

=4m^2-8m+4-4m+12

=4m^2-12m+16

=4m^2-12m+9+7=(2m-3)^2+7>=7>0 với mọi m

=>Phương trình luôn có hai nghiệm phân biệt

\(\left(\dfrac{1}{x1}-\dfrac{1}{x2}\right)^2=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}-\dfrac{2}{x_1x_2}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{\left(\left(x_1+x_2\right)^2-2x_1x_2\right)}{\left(x_1\cdot x_2\right)^2}-\dfrac{2}{x_1\cdot x_2}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{\left(2m-2\right)^2-2\left(m-3\right)}{\left(-m+3\right)^2}-\dfrac{2}{-m+3}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{4m^2-8m+4-2m+6}{\left(m-3\right)^2}+\dfrac{2}{m-3}=\dfrac{\sqrt{11}}{2}\)

=>\(\dfrac{4m^2-10m+10+2m-6}{\left(m-3\right)^2}=\dfrac{\sqrt{11}}{2}\)

=>\(\sqrt{11}\left(m-3\right)^2=2\left(4m^2-8m+4\right)\)

=>\(\sqrt{11}\left(m-3\right)^2=2\left(2m-2\right)^2\)

=>\(\Leftrightarrow\left(\dfrac{m-3}{2m-2}\right)^2=\dfrac{2}{\sqrt{11}}\)

=>\(\left[{}\begin{matrix}\dfrac{m-3}{2m-2}=\sqrt{\dfrac{2}{\sqrt{11}}}\\\dfrac{m-3}{2m-2}=-\sqrt{\dfrac{2}{\sqrt{11}}}\end{matrix}\right.\)

mà m nguyên

nên \(m\in\varnothing\)

1.

đk: \(x\ge2\)

Đặt y = \(\sqrt{x+2}\) ta biến pt về dạng pt thuần nhất bậc 3 đối vs x và y:

ta có : \(x^3-3x^2+2y^3-6x=0\)

\(\Leftrightarrow x^3-3xy^2+2y^3=0\)

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

ta sẽ có nghiệm : \(x=2;x=2-2\sqrt{3}\)

\(1.đk:\left(x+2\right)^3\ge0\Leftrightarrow x\ge-2\)

\(pt\Leftrightarrow x^3-3x\left(x+2\right)+2\sqrt{\left(x+2\right)^3}=0\)

\(\Leftrightarrow x^3-x\left(x+2\right)+2\sqrt{\left(x+3\right)^2}-2x\left(x+2\right)=0\)

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

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

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

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

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

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+2\end{matrix}\right.\)\(\Leftrightarrow x=2\left(tm\right)\)

\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\Leftrightarrow x\le0\\x^2=4\left(x+2\right)\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{3}\left(tm\right)\)

\(2.đk:x^2;y^2\ge2018\Leftrightarrow\left[{}\begin{matrix}x;y\le-\sqrt{2018}\\x;y\ge\sqrt{2018}\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x^2-2018}-\sqrt{y^2-2018}+x^2-y^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)+\dfrac{x^2+11-y^2-11}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{x^2-2018-y^2+2018}{\sqrt{x^2-2018}+\sqrt{y^2-2018}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left[1+\dfrac{1}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{1}{\sqrt{x^2-2018}+\sqrt{y^2+2018}}>0\right]=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(x=y\Rightarrow M=x^{11}-x^{2018}\)

\(x=-y\Rightarrow M=-y^{11}-y^{2018}=:vvv\) (đến đây chịu)