Жақшаларды ашайық:

\(\displaystyle \begin{array}{l}(3xy^{\,2}-2x+4xy\,)+(3xy+3xy^2-xy-11)=\\[10pt]\kern{6em} =3xy^{\,2}-2x+4xy+3x+3xy^{\,2}-xy-11{\small .}\end{array}\)

Ұқсас қосылғыштарды келтірейік:

\(\displaystyle \begin{array}{l}3\color{blue}{xy^{\,2}}-2\color{green}{x}+4\color{red}{xy}+3\color{green}{x}+3\color{blue}{xy^{\,2}}-\color{red}{xy}-11=\\[10pt]\kern{3em} =(3\color{blue}{xy^{\,2}}+3\color{blue}{xy^{\,2}}\,)+(-2\color{green}{x}+3\color{green}{x})+(4\color{red}{xy}-\color{red}{xy}\,)-11=\\[10pt]\kern{7em} =(3+3)\color{blue}{xy^{\,2}}+(-2+3)\color{green}{x}+(4-1)\color{red}{xy}-11=\\[10pt]\kern{19em} =6\color{blue}{xy^{\,2}}+\color{green}{x}+3\color{red}{xy}-11{\small .}\end{array}\)

Жауабы: \(\displaystyle 6{xy^{\,2}}+{x}+3{xy}-11{\small .}\)