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

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

Đặt \(2y=a\)thì ta được

\(P=\frac{1}{x^2+a^2}+\frac{1}{xa}=\left(\frac{1}{x^2+a^2}+\frac{1}{2xa}\right)+\frac{1}{2xa}\)

\(\ge\frac{4}{x^2+a^2+2ax}+\frac{2}{\left(x+a\right)^2}=\frac{6}{\left(x+a\right)^2}\ge\frac{6}{4}=\frac{3}{2}\)

\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

Làm nốt

2/ 

\(T=8x^2-4x+\frac{1}{4x^2}+15\)

\(=\left(4x^2-4x+1\right)+\left(4x^2+\frac{1}{4x^2}-2\right)+16\)

\(=\left(2x-1\right)^2+\left(\frac{4x^2-1}{2x}\right)^2+16\ge16\)

\(Q=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\left(\dfrac{1}{ab}+ab\right)+\dfrac{1}{2ab}\)

\(Q\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{ab}{ab}}+\dfrac{2}{\left(a+b\right)^2}\)

\(Q\ge\dfrac{6}{\left(a+b\right)^2}+2\ge\dfrac{6}{2^2}+2=\dfrac{7}{2}\)

\(Q_{min}=\dfrac{7}{2}\) khi \(a=b=1\)

`Answer:`

Câu 1:

a. \(A=\frac{3}{3-x}-\frac{1}{3+x}-\frac{2x}{9-x^2}\left(ĐKXĐ:x\ne\pm3\right)\)

\(=\frac{3\left(3+x\right)}{\left(3-x\right)\left(3+x\right)}-\frac{1\left(3-x\right)}{\left(3-x\right)\left(3+x\right)}-\frac{2x}{\left(3-x\right)\left(3+x\right)}\)

\(=\frac{9+3x-3+x-2x}{\left(3-x\right)\left(3+x\right)}\)

\(=\frac{6+2x}{\left(3-x\right)\left(3+x\right)}\)

\(=\frac{2}{3-x}\)

b. Ta có: \(A=-\frac{1}{2}\Rightarrow\frac{2}{3-x}=-\frac{1}{2}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow2.2=-1\left(3-x\right)\)

\(\Leftrightarrow4=-3+x\)

\(\Leftrightarrow x=7\)

Thay `x=7` vào `Q`, ta được: \(Q=7^2-7.7+2021=49-49+2021=2021\)

Câu 2:

a. \(\left(x-2\right)\left(2x-1\right)=5\left(2-x\right)\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)-5\left(2-x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1+5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+4\right)=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)

b. \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{x\left(x+5\right)}{x^2-25}\left(ĐK:x\ne\pm5\right)\)

\(\Leftrightarrow\frac{\left(x+5\right)^2-\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x^2+5x}{x^2-25}\)

\(\Leftrightarrow\frac{x^2+10x+25-x^2+10x-25}{x^2-5^2}=\frac{x^2+5x}{x^2-25}\)

\(\Leftrightarrow\frac{20x}{x^2-25}=\frac{x^2+5x}{x^2-25}\)

\(\Leftrightarrow20x=x^2+5x\)

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

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

Câu 3:

\(10\) phút \(=\frac{1}{6}\) giờ

Gọi quãng đường từ nhà đến trường là `x(x>0)`

Thời gian một học sinh đi xe đạp từ nhà đến trường là \(\frac{x}{15}\) giờ

Thời gian một học sinh đi xe đạp từ trường về đến nhà là \(\frac{x}{12}\) giờ

Vì lúc đi về nhà, học sinh đi với vận tốc \(12km/h\) và thời gian về nhiều hơn thời gian đến là \(\frac{1}{6}\) giờ nên ta có phương trình sau:

\(\frac{x}{12}-\frac{x}{15}=\frac{1}{6}\)

\(\Leftrightarrow\frac{15x}{180}-\frac{12x}{180}=\frac{1}{6}\)

\(\Leftrightarrow\frac{3x}{180}=\frac{1}{6}\)

\(\Leftrightarrow18x=180\)

\(\Leftrightarrow x=10km\)

Câu 4:

a. Xét `\triangleABH` và `\triangleACE:`

`\hat{CAE}` chung

`\hat{AHB}=\hat{AEC}=90^o`

`=>\triangleABH` đồng dạng `\triangleACE`

\(\Rightarrow\frac{AB}{AH}=\frac{AC}{AE}\)

\(\Rightarrow AB.AE=AC.AH\)

b. Xét `\triangleCBH` và `\triangleACF:`

`\hat{AFC}=\hat{BHC}=90^o`

`\hat{HCB}=\hat{CAF}`

`=>\triangleCBH` đồng dạng `\triangleACF`

c. Xét `\triangleAHB` và `\triangleCHQ:`

`\hat{AHB}=\hat{HCQ}=90^o`

`\hat{HAB}=\hat{HCQ}`

`=>\triangleAHB` đồng dạng `\triangleCHQ`

\(\Rightarrow\frac{HB}{HQ}=\frac{AH}{HC}\left(1\right)\)

Mà AK//BC \(\Rightarrow\frac{KH}{HB}=\frac{HA}{HC}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrow\frac{HB}{HQ}=\frac{KH}{HB}\)

\(\Rightarrow HB^2=HQ.KH\)

Câu 5:

Đặt \(\hept{\begin{cases}a=2017-x\\b=2019-x\\c=2x-4036\end{cases}}\Rightarrow a+b+c=0\)

Vì \(a+b+c=0\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\)

Ta có \(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab\left(-c\right)+c^3=3abc\)

\(\Rightarrow3\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)

\(\Leftrightarrow2017-x=0\) hoặc \(2019-x=0\) hoặc \(2x-4036=0\)

\(\Leftrightarrow x=2017\) hoặc \(x=2019\) hoặc \(x=2018\)