\(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-\left(x^3+8x^2\right)-27=0\)

\(\Leftrightarrow17x-17=0\\ \Leftrightarrow x=1\)

\(P=\dfrac{mx-2019}{x^2}\Rightarrow px^2-mx+2019=0\)

                            \(\Delta=m^2-4.2019P\ge0\)

                                \(\Leftrightarrow P\le\dfrac{m^x}{8076}\)

để \(\max\limits_P=2019\) thì \(\dfrac{m^2}{8076}=2019\)

                                \(\Leftrightarrow m^2=8076.2019\)

                                \(=2.2.2019.2019\)

                                \(\Leftrightarrow m=4038\)(vì m>0)

vậy m=4038

ÁP dụng tính chất dãy tỉ số bằng nhau:

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

\(\Rightarrow\left\{{}\begin{matrix}z=y\\y=z\\x=z\end{matrix}\right.\)\(\Rightarrow x=y=z=\dfrac{2022}{3}=674\)

\(\Rightarrow\left(x,y,x\right)=674\)

a)\(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2+28x-24-12x^2+15x-3=-27\)

\(\Leftrightarrow43x=0\\ \Leftrightarrow x=0\)

dòng 2 tù dưới lến thiếu bằng nha bấm hơi nhanh

\(A=2x^2\left(3x+4\right)\left(3x-4\right)-\dfrac{9}{2}\left(2x^2+1\right)\left(2x^2-1\right)\)

\(=2x^2\left(9x^2-16\right)-\dfrac{9}{2}\left(4x^4-1\right)\)

\(18x^4-32x^2-18x^4+\dfrac{9}{2}\\ =-32x^2+\dfrac{9}{2}\)

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

\(\Leftrightarrow\left(a+b\right)^2\ge2\\ \Leftrightarrow a+b\ge\sqrt{2}\)

\(\Rightarrow gtnn:\\a+b=\sqrt{2}\)

a)\(A=a\left(a-b\right)-b\left(a-b\right)\)

\(=a^2-ab-ab+b^2\)

\(=a^2-2b+b^2\\ =\left(a-b\right)^2\)

b)\(B=m\left(-2m^2+1\right)+m^2\left(2m^2+1\right)-m\)

\(=-2m^4+m+2m^4+m^3-m\)

\(=m^2\)

c)\(C=\left(-2t\right)^2\left(t+2\right)-8t^2\left(1-t\right)-4t^3\)

\(=4t^2.\left(t+2\right)-8t^2\left(1-t\right)-4t^3\)

\(=4t^3+8t^2-8t^2+8t^3-4t^3\\ =8t^3\)