`@` `\text {Ans}`

`\downarrow`

\(\dfrac{7}{4}x+\dfrac{2}{3}=\dfrac{5}{6}\)

`\Rightarrow`\(\dfrac{7}{4}x=\dfrac{5}{6}-\dfrac{2}{3}\)

`\Rightarrow`\(\dfrac{7}{4}x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div\dfrac{7}{4}\)

`\Rightarrow` `x=2/21`

Vậy, `x = 2/21.`

\(\dfrac{7}{4}.x+\dfrac{2}{3}=\dfrac{5}{6}\)

⇒\(\dfrac{7}{4}.x=\dfrac{5}{6}-\dfrac{2}{3}=\dfrac{1}{6}\)

⇒\(x=\dfrac{1}{6}:\dfrac{7}{4}=\dfrac{1}{6}.\dfrac{4}{7}=\dfrac{2}{21}\)

`@` `\text {Ans}`

`\downarrow`

\(\left(\dfrac{17}{5}+\dfrac{3}{4}\right)\cdot\left(-\dfrac{1}{2}-\dfrac{4}{3}\right)\)

`=`\(\dfrac{83}{20}\cdot\left(-\dfrac{11}{6}\right)=-\dfrac{913}{120}\)

  (\(\dfrac{17}{5}\) + \(\dfrac{3}{4}\)).(-\(\dfrac{1}{2}\) - \(\dfrac{4}{3}\))

=( \(\dfrac{68}{20}\) + \(\dfrac{15}{20}\)).(- \(\dfrac{3}{6}\) - \(\dfrac{8}{6}\))

= \(\dfrac{83}{20}\).(\(-\dfrac{11}{6}\))

= - \(\dfrac{913}{120}\)

Diểm O ở đâu ra vậy em nhỉ, em xem kỹ lại đề bài em nhé!

CCứu em với

Hết cứu:))

\(\left(-\dfrac{3}{4}+\dfrac{5}{13}\right)\times\dfrac{7}{2}-\left(\dfrac{9}{4}-\dfrac{8}{13}\right)\times\dfrac{7}{2}\)

\(=-\dfrac{19}{52}\times\dfrac{7}{2}-\dfrac{85}{52}\times\dfrac{7}{2}\)

\(=\dfrac{7}{2}\times\left(-\dfrac{19}{52}-\dfrac{85}{52}\right)\)

\(=\dfrac{7}{2}\times-2\)

\(=-7\)

\(f\left(x\right)=0\Leftrightarrow x^2+3=0\)

⇔ Vô nghiệm để đa thức f(x)=0 (vì x²≥0⇒x²+3>0)

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

mn ơi giúp mình với mai nộp òi

|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