1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

\(\dfrac{-15}{3}.\dfrac{26}{7}.\dfrac{-21}{5}.\dfrac{15}{16}\)

\(=5.\dfrac{2.13}{7}.\dfrac{3.7}{5}.\dfrac{3.5}{2.8}\)

\(=\dfrac{5.13.3.3}{4}=\dfrac{585}{4}\)

\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)

\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)

\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)

\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)

\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)

\(=-\dfrac{1}{6}.\left(-36\right).\)

\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)

Vậy......

\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)

\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)

\(=\dfrac{5}{8}:\dfrac{15}{16}.\)

\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)

Vậy......

c, (làm tương tự câu b).

~ Học tốt!!! ~

\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)

a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)

\(=\dfrac{2}{3}+\dfrac{2}{7}\)

\(=\dfrac{14}{21}+\dfrac{6}{21}\)

\(=\dfrac{20}{21}\)

\(\dfrac{7}{12}-\dfrac{27}{7}.\dfrac{1}{8}=\dfrac{7}{12}-\dfrac{27}{56}=\dfrac{392}{672}-\dfrac{324}{672}=\dfrac{68}{672}=\dfrac{17}{168}\)

-2/15

-4/35

-5/3

10

-8/3

-7/2

-9

-4/3

Chúc em học giỏi

\(=\dfrac{-2}{15}\\ =\dfrac{-4}{35}\\ =-1\\ =10\\ =\dfrac{-8}{3}\\ =-7\\ =-9\\ =\dfrac{-4}{3}\)

-2/15

-4/35

-5/3

10

-8/3

-7/2

-9

-4/3

a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)

=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)

=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)

=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)

=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)

=\(\dfrac{36}{208}-\dfrac{5}{3}\)

=\(\dfrac{108}{624}-\dfrac{1040}{624}\)

=\(\dfrac{-932}{624}\)

=\(\dfrac{233}{156}\)

                                 còn câu b mk chưa học nên mk chịu

Giải:

5/9:13/7+5/9:13/9 -1 2/3

=5/9.7/13+5/9.9/13-5/3

=5/9.(7/13+9/13)-5/3

=5/9.16/13-5/3

=80/117-5/3

=-115/117

4 2/5 : 0,5% -1 3/7 .14% +(-0,5)

=22/5:1/200-10/7.7/50 +(-1/2)

=880-1/5-1/2

=8793/10

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

a: \(=\dfrac{-8}{9}-\dfrac{6}{5}+\dfrac{8}{9}=-\dfrac{6}{5}\)

c: \(=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}+\dfrac{1}{2}\)

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

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

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

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

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

\(\dfrac{7}{38}-\dfrac{8}{-9}+\dfrac{2}{-9}\)

\(=\dfrac{7}{38}-\dfrac{-8}{9}+\dfrac{-2}{9}\)

\(=\dfrac{7}{38}+\dfrac{8}{9}-\dfrac{2}{9}\)

\(=\dfrac{367}{342}-\dfrac{2}{9}\)

\(=\dfrac{97}{114}\)

Bai de ko bt ngu vai

\(\dfrac{7}{38}\) - \(\dfrac{8}{-9}\) + \(\dfrac{2}{-9}\)

= \(\dfrac{7}{38}\) + \(\dfrac{8}{9}\) - \(\dfrac{2}{9}\)

= \(\dfrac{7}{38}\) + \(\dfrac{6}{9}\)

= \(\dfrac{63}{342}\) + \(\dfrac{228}{342}\)

= \(\dfrac{291}{342}\)

= \(\dfrac{97}{114}\)