1) X2 + 15x + 56 2) X2 – 9x + 20 3) X2 – 6x – 16 4) X2 + x + ¼ 5) 5)x2 – x – 1/4 mấy cái x2 là x mũ 2 nha 01/10/2021 Bởi Jasmine 1) X2 + 15x + 56 2) X2 – 9x + 20 3) X2 – 6x – 16 4) X2 + x + ¼ 5) 5)x2 – x – 1/4 mấy cái x2 là x mũ 2 nha
$$\eqalign{ & 1)\,\,{x^2} + 15x + 56 \cr & = {x^2} + 7x + 8x + 56 \cr & = \left( {{x^2} + 7x} \right) + \left( {8x + 56} \right) \cr & = x\left( {x + 7} \right) + 8\left( {x + 7} \right) \cr & = \left( {x + 7} \right)\left( {x + 8} \right) \cr & 2)\,\,{x^2} – 9x + 20 \cr & = {x^2} – 4x – 5x + 20 \cr & = \left( {{x^2} – 4x} \right) – \left( {5x – 20} \right) \cr & = x\left( {x – 4} \right) – 5\left( {x – 4} \right) \cr & = \left( {x – 4} \right)\left( {x – 5} \right) \cr & 3)\,\,{x^2} – 6x – 16 \cr & = {x^2} – 6x + 9 – 25 \cr & = {\left( {x – 3} \right)^2} – {5^2} \cr & = \left( {x – 3 – 5} \right)\left( {x – 3 + 5} \right) \cr & = \left( {x – 8} \right)\left( {x + 2} \right) \cr & 4)\,\,{x^2} + x + {1 \over 4} \cr & = {x^2} + 2.x.{1 \over 2} + {\left( {{1 \over 2}} \right)^2} \cr & = {\left( {x + {1 \over 2}} \right)^2} \cr & 5)\,\,{x^2} – x – {1 \over 4} \cr & = {x^2} – x + {1 \over 4} – {1 \over 2} \cr & = \left( {{x^2} – 2.x.{1 \over 2} + {{\left( {{1 \over 2}} \right)}^2}} \right) – {\left( {{1 \over {\sqrt 2 }}} \right)^2} \cr & = {\left( {x – {1 \over 2}} \right)^2} – {\left( {{1 \over {\sqrt 2 }}} \right)^2} \cr & = \left( {x – {1 \over 2} – {1 \over {\sqrt 2 }}} \right)\left( {x – {1 \over 2} + {1 \over {\sqrt 2 }}} \right) \cr} $$ Bình luận
$$\eqalign{
& 1)\,\,{x^2} + 15x + 56 \cr
& = {x^2} + 7x + 8x + 56 \cr
& = \left( {{x^2} + 7x} \right) + \left( {8x + 56} \right) \cr
& = x\left( {x + 7} \right) + 8\left( {x + 7} \right) \cr
& = \left( {x + 7} \right)\left( {x + 8} \right) \cr
& 2)\,\,{x^2} – 9x + 20 \cr
& = {x^2} – 4x – 5x + 20 \cr
& = \left( {{x^2} – 4x} \right) – \left( {5x – 20} \right) \cr
& = x\left( {x – 4} \right) – 5\left( {x – 4} \right) \cr
& = \left( {x – 4} \right)\left( {x – 5} \right) \cr
& 3)\,\,{x^2} – 6x – 16 \cr
& = {x^2} – 6x + 9 – 25 \cr
& = {\left( {x – 3} \right)^2} – {5^2} \cr
& = \left( {x – 3 – 5} \right)\left( {x – 3 + 5} \right) \cr
& = \left( {x – 8} \right)\left( {x + 2} \right) \cr
& 4)\,\,{x^2} + x + {1 \over 4} \cr
& = {x^2} + 2.x.{1 \over 2} + {\left( {{1 \over 2}} \right)^2} \cr
& = {\left( {x + {1 \over 2}} \right)^2} \cr
& 5)\,\,{x^2} – x – {1 \over 4} \cr
& = {x^2} – x + {1 \over 4} – {1 \over 2} \cr
& = \left( {{x^2} – 2.x.{1 \over 2} + {{\left( {{1 \over 2}} \right)}^2}} \right) – {\left( {{1 \over {\sqrt 2 }}} \right)^2} \cr
& = {\left( {x – {1 \over 2}} \right)^2} – {\left( {{1 \over {\sqrt 2 }}} \right)^2} \cr
& = \left( {x – {1 \over 2} – {1 \over {\sqrt 2 }}} \right)\left( {x – {1 \over 2} + {1 \over {\sqrt 2 }}} \right) \cr} $$