\(-\frac{3}{10}+x=-\frac{7}{18}\)

\(x=-\frac{7}{10}-\left(-\frac{3}{10}\right)\)

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

\(x=-\frac{2}{5}\)

X=-4/45

bạn mai xuân sai rồi

Ta có:

\(x=\frac{2}{2\sqrt[3]{2}+2\sqrt[3]{4}}=\frac{1}{\sqrt[3]{2}+\sqrt[3]{4}}=\)\(\frac{2\sqrt[3]{2}-2+\sqrt[3]{4}}{6}\)

\(y=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{6\left(\sqrt[3]{2}+\sqrt[3]{4}\right)}{6}\)

\(\Rightarrow xy^3-x^3y=xy\left(y^2-x^2\right)=y^2-x^2=\frac{36\left(\sqrt[3]{4}+4+2\sqrt[3]{2}\right)}{36}\)\(-\frac{4\sqrt[3]{4}+4+2\sqrt[3]{2}-8\sqrt[3]{2}+8-4\sqrt[3]{4}}{36}\)\(=\frac{36\sqrt[3]{4}+144+72\sqrt[3]{2}-12+6\sqrt[3]{2}}{36}=\frac{36\sqrt[3]{4}+78\sqrt[3]{2}+132}{36}\)\(=\frac{6\sqrt[3]{4}+13\sqrt[3]{2}+22}{6}\)

Ta có:

\(S_{ABC}=pr;S_{ACD}=\frac{AC+CD+AD}{2}.r_1;S_{ABD}=\frac{AB+BD+AD}{2}.r_2\)

Vì AD là tia phân giác \(\widehat{BAC}\)nên đường cao từ D đến AB và AC là bằng nhau.

\(\Rightarrow\hept{\begin{cases}S_{ACD}=\frac{S_{ABC}}{3}\\S_{ABD}=\frac{2S_{ABC}}{3}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{AC+CD+AD}{2}.r_1=\frac{pr}{3}\\\frac{AB+BD+AD}{2}.r_2=\frac{2pr}{3}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}AC+CD+AD=\frac{2pr}{3r_1}\left(1\right)\\AB+BD+AD=\frac{4pr}{3r_2}\left(2\right)\end{cases}}\)

Lấy (1) + (2) ta dược

\(AC+CD+AB+BD+2AD=\frac{2pr}{3r_1}+\frac{4pr}{3r_2}\)

\(\Leftrightarrow2p+2AD=\frac{2pr}{3r_1}+\frac{4pr}{3r_2}\)

\(\Leftrightarrow AD=\frac{pr}{3r_1}+\frac{2pr}{3r_2}-p=\frac{pr}{3}\left(\frac{1}{r_1}+\frac{2}{r_2}\right)-p\)

Câu 2 ai vẽ hộ cái hình đi