\(\text{∘ Ans}\)

`A =`\(\left(\dfrac{27}{23}+\dfrac{1}{2}\right)-\left(\dfrac{5}{21}+\dfrac{4}{23}\right)-\dfrac{37}{21}\)

`=`\(\dfrac{27}{23}+\dfrac{1}{2}-\dfrac{5}{21}-\dfrac{4}{23}-\dfrac{37}{21}\)

`=`\(\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(-\dfrac{5}{21}-\dfrac{37}{21}\right)+\dfrac{1}{2}\)

`=`\(\dfrac{23}{23}-\dfrac{42}{21}+\dfrac{1}{2}=1-2+0,5=1-1,5=-0,5\)

-0,5

\(\left(-\dfrac{4}{3}+\dfrac{5}{13}\right):\dfrac{2}{7}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right):\dfrac{2}{7}\\ =\left(-\dfrac{4}{3}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right):\dfrac{2}{7}\\ =-\dfrac{595}{156}:\dfrac{2}{7}\\ =-\dfrac{595}{156}.\dfrac{7}{2}=-\dfrac{4165}{312}\)

\(\dfrac{-4165}{312}\)

Tìm x:

\(\dfrac{2}{3}x-\dfrac{5}{6}-\dfrac{3}{4}=\dfrac{2}{3}\\ \dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{2}{3}+\dfrac{3}{4}\\ \dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{17}{12}\\ \dfrac{2}{3}x=\dfrac{17}{12}+\dfrac{5}{6}\\ \dfrac{2}{3}x=\dfrac{9}{4}\\ x=\dfrac{9}{4}:\dfrac{2}{3}\\ x=\dfrac{27}{8}.\)

1:

a: Xét ΔBHA vuông tại H và ΔBHE vuông tại H có

BH chung

góc ABH=góc EBH

=>ΔBHA=ΔBHE

b: Xét ΔBAD và ΔBED có

BA=BE

góc ABD=góc EBD

BD chung

=>ΔBAD=ΔBED
=>góc BED=góc BAD=90 độ

=>BE vuông góc DE

c: AD=DE

DE<DC

=>AD<DC

(3\(x\) - 2)(\(x+4\)) - (1- \(x\))(2-\(x\)) =(\(x+1\))(\(x-2\))

3\(x^2\) + 12\(x\) - 2\(x\) - 8  - (\(x+1\))(\(x-2\)) -  [-(\(x-2\))](1- \(x\)) = 0 

                    3\(x^2\) + 10\(x\) - 8  -  (\(x-2\))( \(x\) + 1 - 1 + \(x\)) = 0

                    3\(x^2\) + 10\(x\) - 8 - (\(x-2\)). 2\(x\)  = 0

                    3\(x^2\) + 10\(x\) - 8 - 2\(x^2\) + 4\(x\)     = 0

                      \(x^2\) + 14\(x\) -  8 = 0

                    \(x^2\) + 7\(x\) + 7\(x\) + 49 - 57 = 0 

                     \(x\)( \(x\) + 7) + 7(\(x\) + 7) = 57

                       (\(x+7\))(\(x\) + 7) =57

                       (\(x+7\))² = 57

                        \(\left[{}\begin{matrix}x+7=\sqrt{57}\\x+7=-\sqrt{57}\end{matrix}\right.\)

                         \(\left[{}\begin{matrix}x=-7+\sqrt{57}\\x=-7-\sqrt{57}\end{matrix}\right.\)

Vậy \(x\) \(\in\) { -7 - \(\sqrt{57}\); - 7 + \(\sqrt{57}\)}

a) \(\left(2x-3\right)^4=5^4\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)

b) \(\left(2x-3\right)^3=-64\)

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

\(\Rightarrow2x-3=-4\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)

a/

\(\left(2x-3\right)^4=5^4\\ \Leftrightarrow\left|2x-3\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

Vậy...

b/ 2x-3=-64

<=>2x = -61

<=> x = -61/2

Hình thì bn tự vẽ

a, Vì điểm M nằm giữa 2 điểm C và D nên ta có:

  ⇒   CM + MD = CD

Thay số : 1 + MD = 6

              ⇒ MD = 5 cm.

b, Vì I là trung điểm của đoạn MD nên ta có:

   ⇒ MI = ID = MD/2 = 2,5cm

Vậy ID = 2,5 cm.

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