a: \(Q=-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1\)

\(A=x^2y-3x+1-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{2}x^2y-\dfrac{7}{12}xy^2-3x\)

b: \(P=\dfrac{3}{4}xy^2+\dfrac{4}{9}x-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{6}xy^2+\dfrac{16}{9}x-\dfrac{1}{2}x^2y-1\)

a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)

\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)

\(3-a=\dfrac{3}{2}\)

\(a=3-\dfrac{3}{2}\)

\(a=\dfrac{6}{2}-\dfrac{3}{2}\)

\(a=\dfrac{3}{2}\)

b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)

\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)

\(0-a=\dfrac{7}{8}\)

\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)

c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)

\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)

\(\dfrac{8}{6}-a=\dfrac{1}{6}\)

\(a=\dfrac{8}{6}-\dfrac{1}{6}\)

\(a=\dfrac{7}{6}\)

a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)

     a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)

     a  = 2 + 1 - 1 - \(\dfrac{1}{2}\)

     a  = 2 - \(\dfrac{1}{2}\)

     a = \(\dfrac{3}{2}\)

b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)

    (3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)

                     a = - \(\dfrac{7}{8}\)

c,    2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a  = \(\dfrac{1}{6}\)

    a =  2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\)  - \(\dfrac{1}{6}\) 

     a =  (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)

     a = 1 +  \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)

     a = \(\dfrac{7}{6}\)

`#040911`

`a)`

\(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\\ \left(2\dfrac{3}{4}+\dfrac{1}{4}\right)-a=1\dfrac{1}{2}\\ 3-a=1\dfrac{1}{2}\\ a=3-1\dfrac{1}{2}\\ a=\dfrac{3}{2}\\ \text{Vậy, a = }\dfrac{3}{2}\)

`b)`

\(3\dfrac{1}{4}-a-3\dfrac{1}{4}=\dfrac{7}{8}\\ \left(3\dfrac{1}{4}-3\dfrac{1}{4}\right)-a=\dfrac{7}{8}\\0-a=\dfrac{7}{8}\\ a=0-\dfrac{7}{8} \\ a=\dfrac{-7}{8}\)

Bạn xem lại đề, lớp 5 chưa học dấu âm.

`c)`

\(2\dfrac{5}{6}-1\dfrac{1}{2}-a=\dfrac{1}{6}\\ \dfrac{4}{3}-a=\dfrac{1}{6}\\ a=\dfrac{4}{3}-\dfrac{1}{6}\\ a=\dfrac{7}{6}\\ \text{Vậy, a = }\dfrac{7}{6}.\)

a,a+1/4=2 3/4-1 1/2    

a+1/2=5/4

    a=5/4-1/2

     a=3/4

b,a-7/4=13/4-7/9

a-7/4=89/36

        a= 89/36+7/4

         a=152/36

c,3/2-a=17/6-1/6

3/2-a=8/3

       a= 3/2-8/3

       a= -7/6

a: \(1-\left(5\dfrac{4}{9}+a-7\dfrac{7}{18}\right):15\dfrac{3}{4}=0\)

=>\(\left(5+\dfrac{4}{9}+a-7-\dfrac{7}{18}\right):\dfrac{63}{4}=1\)

=>\(\left(a-2+\dfrac{1}{18}\right)=\dfrac{63}{4}\)

=>\(a-\dfrac{35}{18}=\dfrac{63}{4}\)

=>\(a=\dfrac{63}{4}+\dfrac{35}{18}=\dfrac{637}{36}\)

b: \(B=\left(\dfrac{2}{15}+\dfrac{5}{3}-\dfrac{3}{5}\right):\left(4\dfrac{2}{3}-2\dfrac{1}{2}\right)\)

\(=\dfrac{2+5\cdot5-3^2}{15}:\left(4+\dfrac{2}{3}-2-\dfrac{1}{2}\right)\)

\(=\dfrac{2+4^2}{15}:\left(2+\dfrac{2}{3}-\dfrac{1}{2}\right)\)

\(=\dfrac{18}{15}:\dfrac{13}{6}=\dfrac{6}{5}\cdot\dfrac{6}{13}=\dfrac{36}{65}\)

a: \(\dfrac{-1}{2}+\dfrac{2}{3}=\dfrac{-3+4}{6}=\dfrac{1}{6}\)

Số đối là -1/6

b \(-\dfrac{3}{4}+\dfrac{-4}{3}=\dfrac{-9-16}{12}=\dfrac{-25}{12}\)

Số đối là 25/12

c: \(\dfrac{-7}{2}+\dfrac{-3}{4}=\dfrac{-14-3}{4}=\dfrac{-17}{4}\)

Số đối là 17/4

d: \(-2-\dfrac{3}{4}=\dfrac{-8-3}{4}=-\dfrac{11}{4}\)

Số đối là 11/4

\(\Leftrightarrow a^2-\dfrac{4}{3}a-\dfrac{19}{3}=0\)

\(\Leftrightarrow3a^2-4a-19=0\)

\(\text{Δ}=\left(-4\right)^2-4\cdot3\cdot\left(-19\right)=244\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}a_1=\dfrac{4-2\sqrt{61}}{6}=\dfrac{2-\sqrt{61}}{3}\\a_2=\dfrac{2+\sqrt{61}}{3}\end{matrix}\right.\)

\(\dfrac{-4}{3}+a^2-4=\dfrac{7}{3}\) ⇔ \(a^2=\dfrac{4}{3}+4+\dfrac{7}{3}\) ⇔\(a^2=\dfrac{11}{3}+4\) ⇔\(a^2=\dfrac{23}{3}\) ⇔ \(\left[{}\begin{matrix}a=\sqrt{\dfrac{23}{3}}\\a=-\sqrt{\dfrac{23}{3}}\end{matrix}\right.\)

A = 

\(A=1-\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-...+\left(\dfrac{3}{4}\right)^{2020}-\left(\dfrac{3}{4}\right)^{2021}\)

\(\dfrac{3}{4}A=\dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2+\left(\dfrac{3}{4}\right)^3-...+\left(\dfrac{3}{4}\right)^{2021}-\left(\dfrac{3}{4}\right)^{2022}\)

\(\Rightarrow A+\dfrac{3}{4}A=1-\left(\dfrac{3}{4}\right)^{2022}\)

\(\Leftrightarrow\dfrac{7}{4}A=1-\left(\dfrac{3}{4}\right)^{2022}\)

\(\Rightarrow A=\dfrac{4}{7}-\dfrac{4}{7}\left(\dfrac{3}{4}\right)^{2022}\)

Ta có: \(A-\dfrac{4}{7}-\dfrac{4}{7}\left(\dfrac{3}{4}\right)^{2022}< \dfrac{4}{7}< 1\)

\(0< \dfrac{3}{4}< 1\Rightarrow0< \left(\dfrac{3}{4}\right)^{2022}< 1\Rightarrow0< \dfrac{4}{7}\left(\dfrac{3}{4}\right)^{2022}< \dfrac{4}{7}\Rightarrow A>0\)

\(\Rightarrow0< A< 1\Rightarrow\left[A\right]=0\)

a)\(a+a+a+\dfrac{1}{2}x2\dfrac{2}{5}+a+\dfrac{14}{5}+a=134\)

\(5xa+\dfrac{1}{2}x\dfrac{12}{5}+\dfrac{14}{5}=134\)

\(5xa+\dfrac{12}{10}+\dfrac{14}{5}=134\)

\(5xa+\dfrac{6}{5}+\dfrac{14}{5}=134\)

\(5xa+4=134\)

\(5xa=134-4=130\)

\(a=130:5=26\)

b)\(5\dfrac{4}{10}-yx\dfrac{3}{4}=\dfrac{2}{3}\)

\(\dfrac{54}{10}-yx\dfrac{3}{4}=\dfrac{2}{3}\)

\(yx\dfrac{3}{4}=\dfrac{54}{10}-\dfrac{2}{3}\)

\(yx\dfrac{3}{4}=\dfrac{71}{15}\)

\(y=\dfrac{71}{15}:\dfrac{3}{4}\)

\(y=\dfrac{284}{45}\)