Dáp án của mình là f(x)= \sqrt{-3}  −3

\(-x-\dfrac{3}{5}=-\dfrac{6}{7}\)

\(\Rightarrow-\left(x+\dfrac{3}{5}\right)=-\dfrac{6}{7}\)

\(\Rightarrow x+\dfrac{3}{5}=\dfrac{6}{7}\)

\(\Rightarrow x=\dfrac{6}{7}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{9}{35}\)

\(\dfrac{6^3+3\times6^2+3^3}{-13}=\dfrac{2^3\times3^3+3\times2^2\times3^2+3^3}{-13}\)

                             \(=\dfrac{2^3\times3^3+2^2\times3^3+3^3}{-13}=\dfrac{3^3\times\left(2^3+2^2+1\right)}{-13}\) 

                            \(=\dfrac{3^3\times13}{-13}=-9\)

\(\dfrac{6^3+3\cdot6^2+3^3}{-13}\)

\(=\dfrac{216+3\cdot36+27}{-13}\)

\(=\dfrac{216+108+27}{-13}\)

\(=\dfrac{241}{-13}\)

Ta có:
\(\dfrac{998}{555}=1+\dfrac{443}{555}\)
\(\dfrac{999}{556}=1+\dfrac{443}{556}\)
So sánh phân số \(\dfrac{443}{555}\) và \(\dfrac{443}{556}\)
Vì \(555< 556\) nên \(\dfrac{1}{555}>\dfrac{1}{556}\)
\(\Rightarrow1+\dfrac{443}{555}>1+\dfrac{443}{556}\)
Vậy \(\dfrac{998}{555}>\dfrac{999}{556}\)

 Ta có một công thức tổng quát là nếu có phân số \(\dfrac{a}{b}>1\) và \(a,b>0\)thì \(\dfrac{a+1}{b+1}< \dfrac{a}{b}\). Thật vậy, điều này tương đương với \(b\left(a+1\right)< a\left(b+1\right)\Leftrightarrow b< a\), luôn đúng vì \(\dfrac{a}{b}>1\).

 Như vậy, trở lại bài toán, ta thấy \(\dfrac{998}{555}>1\) nên \(\dfrac{999}{556}< \dfrac{998}{555}\).

a có song song vs b.còn vì sao thì tui hog bt.

Nếu \(c\) vuông góc với đường thẳng \(a\) và \(b\) thì 2 góc so le sẽ bằng nhau (\(=90^o\)).

Vậy \(a//b\)

`@` `\text {Ans}`

`\downarrow`

\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^3}\)

`=`\(\dfrac{3^2}{243}\cdot\dfrac{81^2}{3^3}\)

`=`\(\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^3}=\dfrac{1}{3^3}\cdot3^5=\dfrac{3^5}{3^3}=3^2=9\)

\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^2}\)

\(=3^2\cdot\dfrac{1}{3^5}\cdot\left(3^4\right)^2\cdot\dfrac{1}{3^2}\)

\(=3^2\cdot\dfrac{1}{3^5}\cdot3^8\cdot\dfrac{1}{3^2}\)

\(=\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^2}\)

\(=\dfrac{3^2}{3^5}\cdot3^6\)

\(=\dfrac{3^2\cdot3^6}{3^5}\)

\(=3^2\cdot3\)

\(=3^3\)

\(=27\)

\(2^3+3\times\left(\dfrac{1}{2}\right)^0\times\left(\dfrac{1}{2}\right)^2\times4+\left[\left(-2\right)^2\div\dfrac{1}{2}\right]\div8\)

\(=8+3\times1\times\dfrac{1}{4}\times4+\left(4\div\dfrac{1}{2}\right)\div8\)

\(=8+3\times1\times\dfrac{1}{4}\times4+8\div8\)

\(=8+3+1\)

\(=11+1\)

\(=12\)