Cho hai đa thức:

     M(x) = 2x^{3}-9x+5 và N(x) = 2x^{3}+4x^{2}-3

a) Tính M(x) - N(x)            b) Tính N(x) - M(x)

a, Ta có : \(M\left(x\right)-N\left(x\right)=\left(2x^3-9x+5\right)-\left(2x^3+4x^2-3\right)\)

\(=2x^3-9x+5-2x^3-4x^2+3\)

\(=\left(2x^3-2x^3\right)-9x-4x^2+\left(5+3\right)\)

\(=0-4x^2-9x+8=-4x^2-9x+8\)

b, Ta có : \(N\left(x\right)-M\left(x\right)=\left(2x^3+4x^2-3\right)-\left(2x^3-9x+5\right)\)

\(=2x^3+4x^2-3-2x^3+9x-5\)

\(=\left(2x^3-2x^3\right)+4x^2+9x-\left(3+5\right)\)

\(=4x^2+9x-8\)

Đề bài là tính phải không :)

\(\frac{0,4+\frac{2}{7}-\frac{2}{11}}{0,6+\frac{3}{7}-\frac{3}{11}}+\frac{0,25-0,2+\frac{1}{7}}{0,75-0,6+\frac{3}{7}}\)

\(=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}}\)

\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}\right)}\)

\(=\frac{2}{3}+\frac{1}{3}=1\)

|x + 4/15| - |-3,75| = -|- 2,15|

=> |x + 4/15| - 3,75 = 2,15

=> |x + 4/15| = 59/10

=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{59}{10}\\x+\frac{4}{15}=-\frac{59}{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{169}{30}\\x=\frac{-185}{30}\end{cases}}\)

\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\frac{4}{15}\right|=-2,15+3,75\)

\(\left|x+\frac{4}{15}\right|=1,6\)

\(\left|x+\frac{4}{15}\right|=\frac{8}{5}\)

\(\Rightarrow\hept{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{8}{5}-\frac{4}{15}=\frac{4}{3}\\x=\frac{-8}{5}-\frac{4}{15}=\frac{-28}{15}\end{cases}}}\)

\(-2,5+\left|3x+5\right|=-1,5\)

\(\Leftrightarrow\left|3x+5\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}3x+5=1\\3x+5=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=-4\\3x=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=-2\end{cases}}\)

vậy x=-4/3 hoặc x=-2

\(-2,5+\left|3x+5\right|=-1,5\)

                  \(\left|3x+5\right|=-1,5+2,5\)

                  \(\left|3x+5\right|=1\)

Ta có : \(\left|3x+5\right|\ge0\forall x\)

\(\Rightarrow\left|3x+5\right|=3x+5\)

\(\Rightarrow3x+5=1\)

               \(3x=1-5\)

               \(3x=-4\)

                  \(x=-4\div3\)

                  \(x=\frac{-4}{3}\)

\(\left|x-3,5\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}x-3,5=5\\x-3,5=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8,5\\x=-2,5\end{cases}}}\)

|x-3,5|=5

x-3,5=5  hoặc    x-3,5=-5

x=5+3,5  hoặc  x=-5+3,5

x=8,5   hoặc    x=-1,5

vậy x=8,5  hoặc      x=-1,5

\(\left|\frac{-2}{3}+\frac{3}{2}\right|.\left(\frac{-4}{5}-\left|-\frac{3}{-2}-1\right|\right)\)

\(=\left|\frac{5}{6}\right|.\left(\frac{-4}{5}-\left|\frac{1}{2}\right|\right)\)

\(=\frac{5}{6}.\left(\frac{-4}{5}-\frac{1}{2}\right)\)

\(=\frac{5}{6}.\frac{-13}{10}\)

\(=\frac{-13}{12}\)

\(\left|\frac{-2}{3}+\frac{3}{2}\right|.\left(\frac{-4}{5}-\left|-\frac{-3}{2}-1\right|\right)\)

\(=\left|\frac{-4}{6}+\frac{9}{6}\right|.\left(\frac{-4}{5}-\left|\frac{3}{2}-1\right|\right)\)

\(=\left|\frac{5}{6}\right|.\left(\frac{-4}{5}-\left|\frac{3}{2}-\frac{2}{2}\right|\right)\)

\(=\frac{5}{6}.\left(\frac{-4}{5}-\left|\frac{1}{2}\right|\right)\)

\(=\frac{5}{6}.\left(\frac{-4}{5}-\frac{1}{2}\right)\)

\(=\frac{5}{6}.\left(\frac{-8}{10}-\frac{5}{10}\right)\)

\(=\frac{5}{6}.\frac{-13}{10}\)

\(=\frac{5.\left(-13\right)}{6.10}=\frac{-13}{6.2}=\frac{-13}{12}=-1\frac{1}{12}\)

Chúc bạn học tốt

mấy cái dấu kẻ dọc ngắn là giá trị tuyệt đối

\(\left(\frac{-2}{5}+\frac{3}{10}\right)\div\left(\frac{2}{-5}+\left|-\frac{4}{3}-2\right|\right)\)

\(=\frac{-1}{10}\div\left(\frac{-2}{5}+\left|\frac{-10}{3}\right|\right)\)

\(=\frac{-1}{10}\div\left(\frac{-2}{5}+\frac{10}{3}\right)\)

\(=\frac{-1}{10}\div\frac{44}{15}\)

\(=\frac{-3}{88}\)