1 Ko giải phương trình \(x^2+x-3=0\) (1) .Hãy tính \(P=x_1^3-4x_2^2+1019\)

Ta  có \(ac=1\cdot\left(-3\right)< 0\)  \(\Rightarrow pt\) (1) có 2 nghiệm phân biệt \(x_1,x_2\)

Theo định lí Vi-et có: \(\Rightarrow x_1+x_2=-1;x_1x_2=-3\) \(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=-1\\x_1x_2=-3\\x_1+1=-x_2;x_2+1=-x_1\end{matrix}\right.\)

\(\Rightarrow P=x_1^3+1-4x_2^2+4+1014=\left(x_1+1\right)\left(x_1^2-x_1+1\right)-4\left(x_2-1\right)\left(x_2+1\right)+1014=-x_2\left(x_1^2-x_1+1\right)-4\left(x_2-1\right)\left(-x_1\right)+1014\)

\(=-x_1\cdot x_2\cdot x_1+x_1x_2-x_2+4x_1x_2-4x_1+1014=3x_1-3-x_2-12-4x_1+1014=-\left(x_1+x_2\right)+999=1+999=1000\)

Mik ghi nhầm \(\sqrt{x^2+x+1}\) chuyển thành \(\sqrt{x^2+x+2}\) ; chân thành xin lỗi 

Bài 1 GPT: \(x^2+2018\sqrt{2x^2+1}=x+1+2018\sqrt{x^2+x+1}\)(1) ĐKXĐ: \(\forall x\in R\)

(1) \(\Leftrightarrow x^2-x-1+2018\sqrt{2x^2+1}-2018\sqrt{x^2+x+1}=0\)

\(\Rightarrow x^2-x-1+2018\cdot\dfrac{\left(\sqrt{2x^2+1}-\sqrt{x^2+x+2}\right)\left(\sqrt{2x^2+1}+\sqrt{x^2+x+2}\right)}{\sqrt{2x^2+1}+\sqrt{x^2+x+2}}=0\)

\(\Leftrightarrow x^2-x-1+2018\cdot\dfrac{\left(x^2-x-1\right)}{\sqrt{2x^2+1}+\sqrt{x^2+x+2}}=0\)

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

\(\Leftrightarrow x^2-x-1=0\) vì \(1+\dfrac{2018}{\sqrt{2x^2+1}+\sqrt{x^2+x+2}}>1>0\forall x\)

\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{5}{4}=0\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

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

\(\Rightarrow\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\cdot\left(x^2+2x\right)+20=\left(x^2+2x\right)\left(x^2+2x+4\right)+5\left(x^2+2x+4\right)=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)

Áp dụng Bđt Bunhiacopxki vào 2 số \(x^2+4y^2\) và \(1+\dfrac{1}{4}\) có:

\(\left(x^2+4y^2\right)\left(1+\dfrac{1}{4}\right)\ge\left(x+y\right)^2=A^2\Rightarrow A^2\le25\Rightarrow A\le5\)

Dấu = xảy ra \(\Leftrightarrow\dfrac{x^2}{1}=\dfrac{4y^2}{\dfrac{1}{4}}\Leftrightarrow x^2=16y^2\Rightarrow x=4,y=1\) 

\(\dfrac{3}{8}=\dfrac{15}{40},\dfrac{15}{40}>\dfrac{13}{40}\Rightarrow\dfrac{3}{8}>\dfrac{13}{40}\)

Ta cần chứng minh: \(\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\le1\)

\(\Rightarrow a\left(2b+c\right)\left(2c+a\right)+b\left(2a+b\right)\left(2c+a\right)+c\left(2a+b\right)\left(2b+c\right)\le\left(2a+b\right)\left(2b+c\right)\left(2c+a\right)\)\(\Leftrightarrow4abc+2a^2b+2ac^2+a^2c+4abc+2a^2b+2b^2c+ab^2+4abc+2ac^2+2b^2c+bc^2\le9abc+4a^2b+4ac^2+2a^2c+4b^2c+2ab^2+2bc^2\)\(\Leftrightarrow12abc+4a^2b+ab^2+a^2c+4ac^2+4b^2c+bc^2\le9abc+4a^2b+4ac^2+2a^2c+4b^2c+2ab^2+2bc^2\)

\(\Leftrightarrow a^2c+ab^2+bc^2\ge3abc\) 

Áp dụng bđt Cô-si : \(a^2c+ab^2+bc^2\ge3\sqrt[3]{a^2c\cdot ab^2\cdot bc^2}=3abc\)

\(\Rightarrow\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\le1\)

Dấu = xảy ra \(\Leftrightarrow a=b=c\)