Bài 25 trang 12 SGK Toán 8 tập 1

Tính:

a) \({\left( {a + b + c} \right)^2}\);

b) \({\left( {a + b - c} \right)^2}\);

c) \({\left( {a - b - c} \right)^2}\) 

Lời giải

\(\eqalign{
& a)\,\,{\left( {a + b + c} \right)^2} = {\left[ {\left( {a + b} \right) + c} \right]^2} \cr
& = {\left( {a + b} \right)^2} + 2\left( {a + b} \right)c + {c^2} \cr
& = {a^2} + 2ab + {b^2} + 2ac + 2bc + {c^2} \cr
& = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac \cr} \)

\(\eqalign{
& b)\,\,\,{\left( {a + b - c} \right)^2} = {\left[ {\left( {a + b} \right) - c} \right]^2} \cr
& = {\left( {a + b} \right)^2} - 2\left( {a + b} \right)c + {c^2} \cr
& = {a^2} + 2ab + {b^2} + \left( { - 2} \right).ac + \left( { - 2} \right).bc + {c^2}\cr& = {a^2} + 2ab + {b^2} - 2ac - 2bc + {c^2} \cr
& = {a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ac \cr} \)

\(\eqalign{
& c)\,\,{\left( {a - b - c} \right)^2} = {\left[ {\left( {a - b} \right) - c} \right]^2} \cr
& = {\left( {a - b} \right)^2} - 2\left( {a - b} \right)c + {c^2} \cr
& = {a^2} - 2ab + {b^2} + \left( { - 2} \right).ac + \left( { - 2} \right).\left( { - b} \right).c + {c^2} \cr
& = {a^2} - 2ab + {b^2} - 2ac + 2bc + {c^2} \cr
& = {a^2} + {b^2} + {c^2} - 2ab + 2bc - 2ac \cr} \)


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