\(\dfrac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}=1\)

Áp dụng BĐT Bunhiacopxki:

\(\left(\dfrac{2}{3}+1+\dfrac{1}{3}\right)\left(\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}\right)\ge\left(\sqrt{\dfrac{2}{3}.\dfrac{3}{2}x^2}+\sqrt{1.\left(y+\dfrac{z}{2}\right)^2}+\sqrt{\dfrac{1}{3}.\dfrac{3z^2}{4}}\right)^2\)

\(\Leftrightarrow2.1\ge\left(x+y+\dfrac{z}{2}+\dfrac{z}{2}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(\frac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)

\(\Rightarrow\left(x+y+z\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(x+y\le2\Rightarrow-\left(x+y\right)\ge-2\)

Do đó:

\(A=2\left(x+\dfrac{1}{x}\right)+2\left(y+\dfrac{1}{y}\right)-\left(x+y\right)\ge2.2\sqrt{x.\dfrac{1}{x}}+2.2\sqrt{y.\dfrac{1}{y}}-2=6\)

\(A_{min}=6\) khi \(x=y=1\)

Đặt \(x+2021=a\)

\(\Rightarrow2P=2a^2+2\left(a+1\right)^2=4a^2+4a+2=\left(2a+1\right)^2+1\ge1\)

\(\Leftrightarrow P\ge\frac{1}{2}\)

Gọi vận tốc xe đi từ A đến B là x ( x> 0 ) 

xe đi từ B đến A là x - 3 

Theo bài ra ta có pt \(2x+2x-6=46\Leftrightarrow4x=52\Leftrightarrow x=13\left(tm\right)\)

Vậy vận tốc xe đi từ A đến B là 13 km/h 

vận tốc xe đi từ B đến A là 10 km/h 

`Answer:`

a. \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\left(ĐKXĐ:x\ne-1;x\ne-8\right)\)

\(\Leftrightarrow14\left(x+8\right)-14\left(x+1\right)=\left(x+1\right)\left(x+8\right)\)

\(\Leftrightarrow x^2+9x+8=98\)

\(\Leftrightarrow x^2+9x-90=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-15\end{cases}}}\)

b. \(\frac{2x}{x-2}+\frac{7}{x+2}\left(ĐKXĐ:x\ne\pm2\right)\)

\(\Leftrightarrow\frac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{7\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=2\)

\(\Leftrightarrow\frac{2x^2+4x+7x-14}{x^2-4}=2\)

\(\Leftrightarrow2x^2+11x-14=2\left(x^2-4\right)\)

\(\Leftrightarrow2x^2+11x-12=2x^2-8\)

\(\Leftrightarrow11x-14=-8\)

\(\Leftrightarrow11x=6\)

\(\Leftrightarrow x=\frac{6}{11}\)

c. \(\frac{x+1}{2022}+\frac{x+2}{2021}=\frac{x+3}{2020}+\frac{x+4}{2019}\) (Câu này mình sửa lại đề nhé. Vì đề bạn cho sai hoặc thiếu.)

\(\Leftrightarrow\left(\frac{x+1}{2022}+1\right)+\left(\frac{x+2}{2021}+1\right)=\left(\frac{x+3}{2020}+1\right)+\left(\frac{x+4}{2019}+1\right)\)

\(\Leftrightarrow\frac{x+1+2022}{2022}+\frac{x+2+2021}{2021}=\frac{x+3+2020}{2020}+\frac{x+4+2019}{2019}\)

\(\Leftrightarrow\frac{x+2023}{2022}+\frac{x+2023}{2021}-\frac{x+2023}{2020}-\frac{x+2023}{2019}\)

\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2022}+\frac{1}{2021}-\frac{1}{2020}-\frac{1}{2019}\right)=0\)

Do \(\frac{1}{2022}+\frac{1}{2021}-\frac{1}{2020}-\frac{1}{2019}\ne0\)

\(\Rightarrow x+2023=0\Leftrightarrow x=-2023\)