12 + 7 x 8

= 12 + 56

= 68

(8 + 6) x 5

= 14 x 5 

= 70

12+56=68

14x5=70

\(a,\dfrac{6}{11}+6+\dfrac{5}{7}=\dfrac{42+462+55}{77}=\dfrac{559}{77}\)

\(b,\dfrac{9}{8}\times\dfrac{3}{12}:\dfrac{5}{9}=\dfrac{9}{8}\times\dfrac{3}{12}\times\dfrac{9}{5}=\dfrac{243}{480}=\dfrac{81}{160}\)

\(c,\dfrac{8}{7}:4+2=\dfrac{8}{7}\times\dfrac{1}{4}+2=\dfrac{8}{28}+2=\dfrac{2}{7}+2=\dfrac{16}{7}\)

\(d,\dfrac{3}{5}+4:\dfrac{6}{4}=\dfrac{3}{5}+4\times\dfrac{4}{6}=\dfrac{3}{5}+\dfrac{8}{3}=\dfrac{49}{15}\)

ô hoooho bắt lỗi dc câu d nhee, sai đề ròi :)

\(\dfrac{7}{9}+\left(\dfrac{2}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{9}+\left(\dfrac{8}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{9}+\dfrac{1}{4}\)
\(=\dfrac{28}{36}+\dfrac{9}{36}=\dfrac{37}{36}\)
\(-------------\)
\(\dfrac{5}{8}+\dfrac{1}{4}\times\dfrac{2}{3}\)
\(=\dfrac{5}{8}+\dfrac{1}{6}\)
\(=\dfrac{15}{24}+\dfrac{4}{24}=\dfrac{19}{24}\)

\(\dfrac{7}{9}+\left(\dfrac{2}{3}-\dfrac{5}{12}\right)=\dfrac{7}{9}+\dfrac{1}{4}=\dfrac{28}{36}+\dfrac{9}{36}=\dfrac{37}{36}\)

_____________________________________

\(\dfrac{5}{8}+\dfrac{1}{4}\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{1}{6}=\dfrac{15}{24}+\dfrac{4}{24}=\dfrac{19}{24}\)

16𝑥2=3𝑦+143

134.74

31/34

\(125,6+45,7:25\times5\) -> Nhân chia trước cộng trừ sau.

\(=125,6+1,828\times5\)

\(=125,6+9,14\)

\(=134,74\)

= \(\dfrac{31}{34}\)

*) |x-5|=|3x+7|

\(\Leftrightarrow\orbr{\begin{cases}x-5=3x+7\\x-5=-3x-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3x=7+5\\x+3x=-7+5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-2x=12\\4x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-6\\x=\frac{-1}{2}\end{cases}}}\)

thay vào biểu thức A rồi tính

*) |7-2x|=x-3 (x >=3)

\(\orbr{\begin{cases}7-2x=x-3\\-7+2x=x-3\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=-7-3\\2x-x=-3+7\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-10\\x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{10}{3}\\x=4\end{cases}}}\)

thay vào rồi tính biểu thức B

b) \(Q=\dfrac{27-2x}{12-x}=\dfrac{2.\left(12-x\right)+3}{12-x}=2+\dfrac{3}{12-x}\)

Để Q đạt max 

thì \(\dfrac{3}{12-x}\) phải max nên 12 - x phải min và 12 - x > 0 

lại có \(x\inℤ\) 

nên 12 - x = 1 

<=> x = 11 

Khi đó Q = 17

Vậy Q_max = 5 khi x = 11