|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)

|7 - \(\dfrac{3}{4}x\)|  - \(\dfrac{3}{2}\) = 2

|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)

|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)

\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\) 

\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)

 5  - |\(x-3\)| = 5

       |\(x-3\)| = 5 - 5

      |\(x-3\)| = 0

      \(x-3\) = 0 

      \(x\) = 3

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

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

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

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

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

Bài 4

a, \(\dfrac{1}{3.4}\) +  \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{26.27}\)

= \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+ \(\dfrac{1}{26}\) - \(\dfrac{1}{27}\)

= \(\dfrac{1}{3}\) - \(\dfrac{1}{27}\)

= \(\dfrac{8}{27}\)

b,

B =  (\(\dfrac{1}{2}\) - 1).(\(\dfrac{1}{3}\)-1)....(\(\dfrac{1}{2018}\)- 1).(\(\dfrac{1}{2019}\) -1)

B = (-\(\dfrac{1}{2}\))(-\(\dfrac{2}{3}\))....(-\(\dfrac{2017}{2018}\)).(-\(\dfrac{2018}{2019}\))

Xét dãy số: 1; 2; 3; ...; 2018

Dãy số trên có 2018 số hạng 

Vậy B là tích của 2018 phân số âm nên tích B là một số dương

B = \(\dfrac{1}{2019}\)

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

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

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

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

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

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

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

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`

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

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

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

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

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

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

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

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

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

\(=3^2\cdot3\)

\(=3^3\)

\(=27\)

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

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

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

\(=8+3+1\)

\(=11+1\)

\(=12\)

Lời giải:

PT $\Leftrightarrow \frac{2x-1}{203}+1)+(\frac{2x-3}{205}+1)=(\frac{5-2x}{207}-1)-(\frac{2x}{101}+2)+5$

$\Leftrightarrow \frac{2x+202}{203}+\frac{2x+202}{205}=\frac{-(2x+202)}{207}-\frac{2x+202}{101}+5$
$\Leftrightarrow (2x+202)(\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})=5$

$\Leftrightarrow x=\frac{1}{2}[5: (\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})-202]$