\(A=\dfrac{4}{3}-\left(\dfrac{5}{3}-\dfrac{3}{4}+\dfrac{5}{18}\right)+\left(\dfrac{9}{4}-\dfrac{8}{3}\right)\\ A=\dfrac{4}{3}-\dfrac{5}{3}+\dfrac{3}{4}-\dfrac{5}{18}+\dfrac{9}{4}-\dfrac{8}{3}\\ A=\left(\dfrac{4}{3}-\dfrac{5}{3}-\dfrac{8}{3}\right)+\left(\dfrac{3}{4}+\dfrac{9}{4}\right)-\dfrac{5}{18}\\ A=-3+3-\dfrac{5}{18}\\ A=-\dfrac{5}{18.}\)

Lời giải:
Ta thấy:

$\widehat{yBA}+\widehat{BAx}=124^0+56^0=180^0$. Mà 2 góc này ở vị trí trong cùng phía nên $By\parallel Ax$ (đpcm)

\(\dfrac{-4}{7}:x=\dfrac{-2}{5}\)

\(\Rightarrow x=\dfrac{-4}{7}:\dfrac{-2}{5}\)

\(\Rightarrow x=\dfrac{10}{7}\)

\(\dfrac{-4}{7}:x=\dfrac{-2}{5}\)

\(x=\dfrac{-4}{7}.\dfrac{-5}{2}\)

\(x=\dfrac{10}{7}\)

\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)

\(=\dfrac{1}{20}\)

   \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)

= \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)

= \(\dfrac{-4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)

= \(\dfrac{-4}{13}.\dfrac{17}{17}\)

= \(\dfrac{-4}{13}.1\)

= \(\dfrac{-4}{13}\)

= \(\dfrac{-4.5-12.4}{13.17}\)

=\(\dfrac{-4\left(5+12\right)}{13.17}\)

=\(\dfrac{-4.17}{13.17}\)

=\(\dfrac{-4}{13}\)

                      (x + 1)⁴ = (x + 1)³

⇒    (x + 1)⁴ - (x + 1)³ = 0

⇒ (x + 1)³ . (x + 1 - 1) = 0

⇒               (x + 1)³ . x = 0

⇒ \(\left[{}\begin{matrix}\left(x+1\right)^3=0\\x=0\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

   Vậy \(x\in\left\{0;-1\right\}\)

`(x+1)^4 =(x+1)^3`

`@TH1: x+1=0 =>x=-1`

    `=>(-1)^4 = (-1)^3`

    `=>1=-1` (Vô lí) 

 `=>x=-1` loại

`@TH2: x+1`\(\ne 0<=>x \ne -1\)

   `=>x+1=1`

   `=>x=0` (t/m)

Vậy `x=0`

\(B=10-5x^2+6x+5x^2-6x=10\) không đổi

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