Answer:

\(\left(5x-3\right)\left(5x+3\right)-\left(5x-2\right)^2\)

\(=\left(25x^2-9\right)-\left(25x^2-20x+4\right)\)

\(=25x^2-9-25x^2+20x-4\)

\(=20x-13\)

\(\left(x-1\right)\left(2x-3\right)+\left(x-2\right)\left(1-2x\right)\)

\(=\left(2x^2-3x-2x+3\right)+\left(x-2x^2-2+4x\right)\)

\(=\left(2x^2-2x^2\right)-\left(5x+5x\right)+\left(3-2\right)\)

\(=1\)

Bài 2:

\(4x^2-1=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\)

\(3x\left(x-2\right)-2\left(2-x\right)=0\)

\(\Rightarrow3x\left(x-2\right)+2\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)

\(9x^2-4-2\left(3x-2\right)=0\)

\(\Rightarrow9x^2-4-6x+4=0\)

\(\Rightarrow9x^2-6x=0\)

\(\Rightarrow x\left(9x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\9x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)

\(3x^2-7x+4=0\)

\(\Rightarrow\left(3x^2-4x\right)+\left(-3x+4\right)=0\)

\(\Rightarrow x\left(3x-4\right)-\left(3x-4\right)=0\)

\(\Rightarrow\left(3x-4\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}}\)

what 

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< đúng ko hả quá dễ

hahaha

199,108,99,8.

Answer:

\(A=\frac{3x}{x+2}+\frac{6}{x+2}\) \(\left(ĐK:x\ne-2\right)\)

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

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

\(=3\)

\(B=\frac{x+1}{x-2}+\frac{x}{x+2}\) \(\left(ĐKXĐ:x\ne\pm2\right)\)

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

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

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

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

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

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

\(C=\frac{2x-1}{x-3}+\frac{x}{x+3}+\frac{5-x-3x^2}{x^2-9}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)

\(=\frac{\left(2x-1\right)\left(x+3\right)+x\left(x-3\right)+5-x-3x^2}{\left(x-3\right)\left(x+3\right)}\)

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

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

theo công thức tính số đường chéo của 1 đa giác lồi có n cạnh = n.(n-3)/2          (n>=3)

---> đa giác lồi có 20 đường chéo thì có số cạnh là :

n.(n-3)/2 = 20

---> n² - 3n -40 = 0

---->n² - 8n + 5n - 40 = 0

---->n.(n-8) + 5.(n-8)

---->(n+5) . (n-8) = 0

-----> +)n = -5

          +)n = 8

mà n>=3

-----------> n = 8

Vậy đa giác lồi có 20 đường chéo thì có 8 cạnh.

tham khảo:https://lazi.vn/edu/exercise/925105/cho-tu-giac-abcd-co-2-duong-cheo-ac-va-bd-vuong-goc-vvoi-nhau-va-cat-nhau-tai-o-tu-o-ke-ok-vvuong-goc-dc-ok-kkeo-dai-cat-ab-o-i-ch