Bn cần bài nào trong 2 bài nhỉ?

e tách câu hỏi ra nhe tạm thời cj giúp mụt câu nhe

a: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=\dfrac{-5}{7}+1+\dfrac{5}{7}=1\)

b: \(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}\cdot4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{13}{56}\)

c: \(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{-24+45}{54}\cdot\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{21}{54}\cdot\dfrac{12}{7}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}\)

d: \(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{108}{333}=\dfrac{12}{37}\)

mik làm vậy đúng ko

\(c,\Leftrightarrow3x-5-x-7=2\Leftrightarrow2x-12=2\Leftrightarrow2x=14\Leftrightarrow x=7\)

\(d,\left\{{}\begin{matrix}x=6\\y-1=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=7\end{matrix}\right.\)

Câu C:

(3x−5)−(x+7)=2

3x−5−x−7=2

2x−5−7=2

2x−12=2

2x=2+12

2x=14

x=214​

x=7

Ban can giup bai nao a

bn o dau minh moi biet de lam