\(2\cdot x-25=125\)

           \(2x=125+25\)

            \(2x=150\)

              \(x=150:2\)

              \(x=75.\)

\(Vậy...\)

\(120-3\cdot x=0\)

          \(-3x=0-120\)

          \(-3x=-120\)

               \(x=\left(-120\right):\left(-3\right)\)

              \(x=40.\)

\(Vậy...\)

\(2\cdot x-11=23\)

           \(2x=23+11\)

           \(2x=34\)

             \(x=34:2\)

             \(x=17.\)

\(Vậy...\)

\(a.M=\left(2a+b\right)^2-\left(b-2a\right)^2\)

\(=4a^2+4ab+b^2-b^2+4ab-4a^2=4a^2-4a^2+4ab+4ab+b^2-b^2\)

\(=8ab.\)

\(b.N=\left(3a+2\right)^2+2\left(2+3a\right)\left(1-2b\right)+\left(2b-1\right)^2\)

\(=9a^2+12a+4+4-8a+6a-12ab+4b^2-4b+1\)

\(=9a^2+12a+6a+4b^2-8b-4b-12ab+4+4+1\)

\(=9a^2+18a+4b^2-12b-12ab+9.\)

\(a.\dfrac{-3}{-7}\cdot\dfrac{14}{5}=\dfrac{3}{7}\cdot\dfrac{14}{5}=\dfrac{42}{35}=\dfrac{6}{5}.\)

\(b.\dfrac{28}{17}\cdot\dfrac{68}{-14}=\dfrac{-28}{17}\cdot\dfrac{34}{7}=-8.\)

\(c.\dfrac{-35}{46}\cdot\dfrac{23}{-205}=\dfrac{35}{46}\cdot\dfrac{23}{205}=\dfrac{7}{82}.\)

\(d.\dfrac{8}{17}\cdot\left(\dfrac{-68}{4}\right)=\dfrac{8}{17}\cdot17=-\dfrac{8}{1}=-8.\)

\(\left(-312\right)+\left(-327\right)+\left(-28\right)+27\)

\(=-639-28+27=-667+27=-640.\)

lx

\(k.x-\dfrac{8}{17}\cdot\left(-\dfrac{68}{5}\right)=\dfrac{3}{7}\)

            \(x+\dfrac{8}{17}\cdot\dfrac{68}{5}=\dfrac{3}{7}\)

                   \(x+\dfrac{32}{5}=\dfrac{3}{7}\)

                             \(x=\dfrac{3}{7}-\dfrac{32}{5}\)

                             \(x=-\dfrac{209}{35}.\)

\(Vậy...\)

\(l.-\dfrac{5}{6}\cdot\dfrac{78}{25}+x=\dfrac{4}{15}\cdot\dfrac{-3}{8}\)

        \(\dfrac{-13}{5}+x=\dfrac{1}{15}\cdot\dfrac{-3}{2}\)

        \(-\dfrac{13}{5}+x=-\dfrac{3}{30}\)

        \(-\dfrac{13}{5}+x=-\dfrac{1}{10}\)

                     \(x=-\dfrac{1}{10}+\dfrac{13}{5}\)

                     \(x=\dfrac{5}{2}.\)

\(Vậy...\)

\(m.21\cdot\dfrac{33}{14}-x=-\dfrac{16}{2}\)

          \(\dfrac{99}{2}-x=-\dfrac{16}{2}\)

          \(\dfrac{99}{2}-x=-8\)

                 \(-x=-8-\dfrac{99}{2}\)

                 \(-x=-\dfrac{115}{2}\)

                    \(x=\dfrac{115}{2}.\)

\(Vậy...\)

\(n.\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}-x=1\)

                    \(-x+\dfrac{7}{9}=1\)

                            \(-x=1-\dfrac{7}{9}\)

                            \(-x=\dfrac{2}{9}\)

                               \(x=\dfrac{2}{9}\)

\(Vậy...\)

cái đề khó hiểu và nó lộn xộn quá bạn ơi.

Bạn làm ơn viết lại nó dùm với ạ.

Vào cái nút M bị quay sang một bên ấy ạ.

lx

\(\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}-\dfrac{4}{3-\sqrt{5}}\)

\(=\dfrac{3\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\dfrac{3\sqrt{5}+3\sqrt{2}-5-\sqrt{10}}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\dfrac{4\left(\sqrt{10}+2-\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)

\(=\dfrac{16\sqrt{2}-8\sqrt{10}-22+10\sqrt{5}}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\dfrac{16\sqrt{2}-8\sqrt{10}-22+10\sqrt{5}}{4\sqrt{10}-8\sqrt{2}+11-5\sqrt{5}}\)

\(=-2.\)

\(\dfrac{6-6\sqrt{3}}{1-\sqrt{3}}+\dfrac{3\sqrt{3}+3}{\sqrt{3}+3}=6+\sqrt{3}.\)