ĐỀ bÀI \(\Leftrightarrow\left(x+\frac{1}{x}\right)^2-2m\left(x+\frac{1}{x}\right)-1=0\left(1\right)\)

đặt \(t=x+\frac{1}{x},\left|t\right|\ge2\)

ta có \(t^2-2mt-1=0\left(2\right)\)

PT(2) luôn có 2 nghiệm \(t_1< 0< t_2\)=> PT (1) có nghiệm khi zà chỉ khi PT(2) có ít nhất 1 nghiệm t sao cho \(\left|t\right|\ge2\)

hay ít nhất 2 số 2 zà -2 phải nằm giữa 2 nghiệm (t1) zà (t2) 

hay \(\orbr{\begin{cases}f\left(2\right)\le0\\f\left(-2\right)\le0\end{cases}=>\orbr{\begin{cases}3-4m\le0\\3+4m\le0\end{cases}=>\orbr{\begin{cases}m\ge\frac{3}{4}\\m\le-\frac{3}{4}\end{cases}}}}\)

#Quá Khứ . IS ! 

\(\sqrt{2x^2-2x+3}=3-x\)

\(\Leftrightarrow2x^2-2x+3=\left(3-x\right)^2\)(\(x\le3\))

\(\Leftrightarrow2x^2-2x+3=9-6x+x^2\)

\(\Leftrightarrow x^2+4x-6=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{10}\left(tm\right)\\x=-2-\sqrt{10}\left(tm\right)\end{cases}}\)

ĐK: \(x\ge\frac{3}{2}\)

Ta có: \(\left(5x-4\right)\sqrt{2x-3}-\left(4x-5\right)\sqrt{3x-2}=2\)

\(\Leftrightarrow\left(5x-4\right)\sqrt{2x-3}=2+\left(4x-5\right)\sqrt{3x-2}\)

\(\Leftrightarrow50x^3-155x^2+152x-48=48x^3-152x^2+155x-46+4\left(4x-5\right)\sqrt{3x-2}\)

\(\Leftrightarrow2x^3-3x^2-3x-2-4\left(4x-5\right)\sqrt{3x-2}=0\)

\(\Leftrightarrow\left(4x-5\right)\sqrt{3x-2}\left(\sqrt{3x-2}-4\right)+2x^3-15x^2+20x-12=0\)

\(\Leftrightarrow\frac{3\left(x-6\right)\left(4x-5\right)\sqrt{3x-2}}{\sqrt{3x-2}+4}+\left(x-6\right)\left(2x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left[\frac{3\left(4x-5\right)\sqrt{3x-2}}{\sqrt{3x-2}+4}+2x^2-3x+2\right]=0\Leftrightarrow x=6\)

Vì \(\frac{3\left(4x-5\right)\sqrt{3x-2}}{\sqrt{3x-2}+4}+2x^2-3x+2>0,\forall x\ge\frac{3}{2}\)

Vậy  pt có nghiệm duy nhất x=6

Đặt \(a=\sqrt[3]{7-x},b=\sqrt[3]{x-5}\Rightarrow a^3+b^3=2,a^3-b^3=12-2x\)

Ta có hệ: 

\(\hept{\begin{cases}\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\\a^3+b^3=2\end{cases}}\Rightarrow\frac{a-b}{a+b}=\frac{a^3-b^3}{a^3+b^3}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)\left(a^2-ab+b^2\right)}\Rightarrow a^2+ab+b^2=a^2-ab+b^2\)

\(\Rightarrow ab=0\)\(\Leftrightarrow\orbr{\begin{cases}a=0\\b=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=7\end{cases}}}\)(thử lại thỏa mãn).

Lấy điểm \(I\)thỏa mãn: \(\overrightarrow{IA}+\overrightarrow{IB}+5\overrightarrow{IC}=\overrightarrow{0}\). 

\(MA^2+MB^2+5MC^2\)

\(=\overrightarrow{MA}^2+\overrightarrow{MB}^2+5\overrightarrow{MC}^2\)

\(=\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IB}\right)^2+5\left(\overrightarrow{MI}+\overrightarrow{IC}\right)^2\)

\(=7MI^2+IA^2+IB^2+5IC^2=4a^2\)

\(\Rightarrow MI=\sqrt{\frac{4a^2-IA^2-IB^2-5IC^2}{7}}=r\)

Suy ra \(M\)thuộc đường tròn tâm \(I\)bán kính \(r\).

Tính tan15^o

\(\cos^215^o=1-\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^2=\frac{8+2\sqrt{12}}{16}\)

\(=\frac{\left(\sqrt{6}\right)^2+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^2}{16}=\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{16}\)

Vì 15^o<90^o nên cos15^o>0 => cos15^o=\(\frac{\sqrt{6}+\sqrt{2}}{4}\)

tan 15^o \(=\frac{sin15^0}{c\text{os}15^0}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)}{6-2}\right)^2=2-\sqrt{3}\)

CM:

\(2\sin15^0\cos15^0=2\frac{\sqrt{6}-\sqrt{2}}{4}.\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{1}{2}=\sin30^0\)

BĐT cần chứng minh tương đương với: 

\(\frac{a}{b}-\frac{a}{b+c}+\frac{b}{c}-\frac{b}{c+a}+\frac{c}{a}-\frac{c}{a+b}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{ac}{b\left(b+c\right)}+\frac{ba}{c\left(c+a\right)}+\frac{cb}{a\left(a+b\right)}\ge\frac{3}{2}\)

Ta có: 

\(\frac{ac}{b\left(b+c\right)}+\frac{ba}{c\left(c+a\right)}+\frac{cb}{a\left(a+b\right)}\)

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

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

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

Bất đẳng thức cần chứng minh sẽ đúng nếu ta chứng minh được \(\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c\right)}\ge3\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)

Đặt \(ab=x,bc=y,ca=z\)suy ra ta cần chứng minh 

\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx\ge3xy+3yz+3zx\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)(đúng) 

Vậy bất đẳng thức ban đầu là đúng. 

Dấu \(=\)xảy ra khi \(a=b=c\).