Tổng các hệ số của đa thức đã cho sau khi khai triển là: 

\(A\left(1\right)=\left(3-4.1+1^2\right)^{2004}.\left(3+4.1+1^2\right)^{2005}=0\)

CÂU TRA LỜI LÀ D

Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Leftrightarrow a=b=c\).

Khi đó \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{2a}{a}+\frac{2a}{a}+\frac{2a}{a}=2+2+2=6\).

Chọn C. 

Bài 1 :

a) \(-\frac{3}{4}.31\frac{11}{23}-0,75.8\frac{12}{23}\)

\(=\left(-1\right).\frac{3}{4}.31\frac{11}{23}-\frac{3}{4}.8\frac{12}{23}\)

\(=\frac{3}{4}.\left[\left(-1\right).31\frac{11}{23}-8\frac{12}{23}\right]\)

\(=\frac{3}{4}.\left(-31\frac{11}{23}-8\frac{12}{23}\right)\)

\(=\frac{3}{4}.\left(-40\right)=-30\)

b) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+5\frac{4}{9}:\left(-\frac{5}{7}\right)\)

\(=4\frac{5}{9}.\frac{7}{5}+5\frac{4}{9}.\frac{7}{5}\)

\(=\left(4\frac{5}{9}+5\frac{4}{9}\right).\frac{7}{5}\)

\(=10.\frac{7}{5}=14\)

c) \(\left(\frac{12}{35}-\frac{6}{7}+\frac{18}{14}\right):\frac{6}{-7}-\frac{-2}{5}-\left(-2021\right)^0\)

\(=\left(\frac{12}{35}-\frac{30}{35}+\frac{45}{35}\right):\frac{6}{-7}+\frac{2}{5}-1\)

\(=27.\frac{-7}{6}-\frac{3}{5}\)

\(=\frac{-321}{10}\)

Bài 2 :

a) \(\left(2x-3\right)\left(\frac{3}{4}x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\\frac{3}{4}x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=3\\\frac{3}{4}x=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-4}{3}\end{cases}}\)

b) \(\frac{2}{3}x+\left(\frac{5}{7}\right)^0=\frac{3}{10}\)

\(\Rightarrow\frac{2}{3}x+1=\frac{3}{10}\)

\(\Rightarrow\frac{2}{3}x=\frac{-7}{10}\)

\(\Rightarrow x=\frac{-7}{10}:\frac{2}{3}=\frac{-21}{20}\)

c) \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)

\(\Rightarrow\frac{1}{7}:x=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)

\(\Rightarrow x=\frac{1}{7}:\frac{-3}{14}=\frac{-2}{3}\)

d) \(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)

\(\Rightarrow\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=-3+1+1+1\)

\(\Rightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)

\(\Rightarrow\left(x+2010\right).\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

\(\Rightarrow x+2010=0\left(\text{ do }\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\ne0\right)\)

=> x = -2010

câu c nha

nhớ tít nha

da em thua chi em la hoc sinh lop 2  a

ko sao đâu

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\)

\(\Rightarrow\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=6\)