\(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\)

\(x+\frac{1}{8}=\frac{1}{4}\)

\(x=\frac{1}{4}-\frac{1}{8}\)

\(x=\frac{4}{16}-\frac{2}{16}\)

\(x=\frac{1}{8}\)

Vậy \(x=\frac{1}{8}\)

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

      \(\frac{8}{27}-x=\frac{1}{3}\)

                    \(x=\frac{8}{27}-\frac{1}{3}\)

                    \(x=\frac{8}{27}-\frac{9}{27}\)

                     \(x=-\frac{1}{27}\)

Vậy \(x=-\frac{1}{27}\)

c) \(x.\left(-\frac{1}{2}\right)^4=\frac{3}{8}\)

 \(x.\frac{1}{16}=\frac{3}{8}\)

       \(x=\frac{3}{8}:\frac{1}{16}\)

        \(x=\frac{3}{8}.16\)

      \(x=6\)

c) \(\left(\frac{1}{2}\right)^3.x=\left(\frac{1}{2}\right)^5\)

\(x=\left(\frac{1}{2}\right)^5:\left(\frac{1}{2}\right)^3\)

\(x=\left(\frac{1}{2}\right)^2\)

\(x=\frac{1}{4}\)

Vậy \(x=\frac{1}{4}\)

Chúc bạn học tốt !!!

a) \(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\Leftrightarrow x+\frac{1}{8}=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}-\frac{1}{8}\Leftrightarrow x=\frac{1}{8}\)

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

c) \(x.\left(\frac{-1}{2}\right)^4=\frac{3}{8}\Leftrightarrow x.\frac{1}{16}=\frac{3}{8}\Leftrightarrow x=\frac{3}{8}:\frac{1}{16}\Leftrightarrow x=6\)

d) \(\left(\frac{1}{2}\right)^2.x=\left(\frac{1}{2}\right)^5\Leftrightarrow\frac{1}{8}.x=\frac{1}{32}\Leftrightarrow x=\frac{1}{32}:\frac{1}{8}\Leftrightarrow x=\frac{1}{4}\)

(x-3²)=81

suy ra x-3²=81

suy ra x=81+9

suy ra x=90

\(x-3^2=81\)

\(\Leftrightarrow x-9=81\)

\(\Leftrightarrow x=81+9\)

\(\Leftrightarrow x=90\)

\(=\)\(\frac{1}{2}\)-\(\frac{-13}{6}\)-\(\frac{5}{6}\)

\(=\)\(\frac{3}{6}\)-\(\frac{-13}{6}\)-\(\frac{5}{6}\)

\(=\)\(\frac{11}{6}\)

\(\frac{3}{4}-0,25-\left(\frac{7}{3}+\frac{-9}{2}\right)-\frac{5}{6}\)

\(=\frac{3}{4}-\frac{1}{4}-\left(\frac{14}{6}+\frac{-27}{6}\right)-\frac{5}{6}\)

\(=\frac{1}{2}-\frac{-13}{6}-\frac{5}{6}\)

\(=\frac{3}{6}+\frac{13}{6}-\frac{5}{6}\)

\(=\frac{16}{6}-\frac{5}{6}\)

\(=\frac{11}{6}\)

a) Vì AB = AC mà E là trung điểm của AB, F là trung điểm của AC 

\(\Rightarrow\) AE = AF

(Tự vẽ hình)

a) Ta có: E là trung điểm của AB => AE=EB=1/2AB

              F là trung điểm của AC => AF=FC=1/2AC

Mà AB=AC nên AE=AF

b) Xét tg ABF và tg ACE có:

AB=AC ( Giả thiết )

\(\widehat{A}\): chung

AE=AF ( cmt câu a)

Do đó: tg ABF=tgACE ( c_g_c)

=>\(\widehat{ABF}=\widehat{ACE}\)( 2 cạnh tương ứng )

Học tốt!

\(A=\left(\frac{3}{5}-0.435+\frac{1}{200}\right)\div\frac{\left(-0.04\right)}{30.75}+\frac{1}{12}+3\frac{1}{6}\)

\(A=\left(\frac{3}{5}-\frac{87}{200}+\frac{1}{200}\right)\div\frac{-4}{3075}+\frac{1}{12}+\frac{19}{6}\)

\(A=\left(\frac{120}{200}-\frac{87}{200}+\frac{1}{200}\right)\div\frac{-4}{3075}+\frac{1}{12}+\frac{19}{6}\)

\(A=\frac{34}{200}×\frac{3075}{-4}+\frac{1}{12}+\frac{19}{6}\)

\(A=\frac{-2091}{16}+\frac{1}{12}+\frac{19}{6}\)

\(A=\frac{-6273}{48}+\frac{4}{48}+\frac{152}{48}\)

\(A=\frac{-6117}{48}=\frac{-2039}{16}\)

Cbht

ban oi phan tat ca ma

a) Ta có: \(A=5m\left(x^2y^3\right)^3=5mx^6y^9\)

Vậy: A và B đồng dạng

b) \(A-B=5mx^6y^9-\left(-\frac{2}{m}\right)x^6y^9\)

\(=\left(5-\frac{2}{m}\right)x^6y^6=\frac{5m^2+2}{m}x^2y^9\)

kết quả là bao nhiêu bạn