\(1,\)

\(A=-\frac{7}{12}+\frac{12}{18}+\frac{5}{4}\)

\(=-\frac{7}{12}+\frac{2}{3}+\frac{5}{4}\)

\(=-\frac{7}{12}+\frac{8}{12}+\frac{15}{12}\)

\(=\frac{-7+8+15}{12}\)

\(=\frac{4}{3}\)

\(1,\)

\(B=\frac{1}{4}-\frac{8}{7}:8-3:\frac{3}{4}.\left(-2\right)^2\)

\(=\frac{1}{4}-\frac{8}{7}.\frac{1}{8}-3.\frac{4}{3}.4\)

\(=\frac{1}{4}-\frac{1}{7}-16\)

\(=\frac{7-4-448}{28}\)

\(=-\frac{445}{28}\)

Mình đang bận, bạn cần gấp thế à? Trả lời trong tin nhắn nhé!!!

2.B=1/4-(8/7:8)-(3:3/8).(-2)^2=1/4-(8/7.1/8)-(3/1.4/3).(-2)^2=1/4-1/7-4.(-2)^2=1/4-1/7-16= 7/28-4/28-448/28= -445/28

\(\frac{4-x}{x+5}=\left(-\frac{2}{3}\right)^2\)

=> \(\frac{4-x}{x+5}=\frac{4}{9}\)

=> 9.(4-x)=4.(x+5)

=> 36-9x=4x+20

=> -9x-4x=20-36

=> -13x=-16

=> x=\(\frac{-16}{-13}=\frac{16}{13}\)

a, \(Chof\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

- Lập bảng xét dấu :

Vậy \(\left\{{}\begin{matrix}f\left(x\right)>0\Leftrightarrow x\in\left(3;4\right)\\f\left(x\right)< 0\Leftrightarrow x\in\left(-\infty;3\right)\cup\left(4;+\infty\right)\\f\left(x\right)=0\Leftrightarrow x\in\left\{3;4\right\}\end{matrix}\right.\)

b, \(f\left(x\right)=\left(x-1\right)\left(x+6\right)\)

( Làm tương tự câu a )

a: \(\Leftrightarrow x^2+10x+25-x^2+4x=55\)

=>14x=30

hay x=15/7

b: \(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\)

hay \(x\in\left\{7;3\right\}\)

chắc b