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Question

Tìm GTNN:

a)                 $\dfrac{x^2+x+1}{x^2+2x+1}$

b)  $\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}$

c) $\dfrac{x^2-4x+1}{x^2+9}$

Trần Quốc Lộc · Accepted Answer

\(\text{a)           }\dfrac{x^2+x+1}{x^2+2x+1}\
=\dfrac{x^2+2x-x+1+1-1}{x^2+2x+1}\
=\dfrac{\left(x^2+2x+1\right)-\left(x+1\right)+1}{x^2+2x+1}\ =\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{x+1}{\left(x+1\right)^2}+\dfrac{1}{\left(x+1\right)^2}\
=1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\left(1\right)\
Đặt\text{                        }\dfrac{1}{x+1}=y\
\Rightarrow\left(1\right)=1-y+y^2\
=y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\
=\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\
Do\text{                  }\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\
\Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\
Dấu\text{                  }"="\text{                  }xảy\text{                  }ra\text{                  }khi:\
\left(y-\dfrac{1}{2}\right)^2=0\
\Leftrightarrow y-\dfrac{1}{2}=0\
\Leftrightarrow y=\dfrac{1}{2}\ \Leftrightarrow\dfrac{
1}{x+1}=\dfrac{1}{2}\
\Leftrightarrow x+1=2\
\Leftrightarrow x=1\
Vậy\text{                  }GTNN\text{                  }của\text{                  }phân\text{                  }thức\text{                  }là\text{                  }\dfrac{3}{4}\text{                  }khi\text{                  }x=1\)

\(\text{b)                      }\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\ =\dfrac{4x^2-4x-2x+1+1-1}{\left(2x-1\right)^2}\ =\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\ =1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\left(1\right)\ Đặt\text{                  }-\dfrac{1}{2x-1}=y\ \Rightarrow\left(1\right)=1+y+y^2\ =y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\ =\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\ =\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ Do\text{                  }\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\ \Rightarrow\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\ Dấu\text{                  }"="\text{                  }xảy\text{                  }ra\text{                  }khi:\
\left(y+\dfrac{1}{2}\right)^2=0\
\Leftrightarrow y+\dfrac{1}{2}=0\
\Leftrightarrow y=-\dfrac{1}{2}\
\Leftrightarrow-\dfrac{1}{2x-1}=-\dfrac{1}{2}\
\Leftrightarrow2x-1=2\
\Leftrightarrow2x=3\
\Leftrightarrow x=\dfrac{3}{2}\
Vậy\text{                     }GTNN\text{                     }của\text{                     }biểu\text{                     }thức\text{                     }là\text{                     }\dfrac{3}{4}\text{                     }khi\text{                     }x=\dfrac{3}{2}\)Câu trả lời: \(\text{a)           }\dfrac{x^2+x+1}{x^2+2x+1}\
=\dfrac{x^2+2x-x+1+1-1}{x^2+2x+1}\
=\dfrac{\left(x^2+2x+1\right)-\left(x+1\right)+1}{x^2+2x+1}\ =\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{x+1}{\left(x+1\right)^2}+\dfrac{1}{\left(x+1\right)^2}\
=1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\left(1\right)\
Đặt\text{                        }\dfrac{1}{x+1}=y\
\Rightarrow\left(1\right)=1-y+y^2\
=y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\
=\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\
Do\text{                  }\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\
\Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\
Dấu\text{                  }"="\text{                  }xảy\text{                  }ra\text{                  }khi:\
\left(y-\dfrac{1}{2}\right)^2=0\
\Leftrightarrow y-\dfrac{1}{2}=0\
\Leftrightarrow y=\dfrac{1}{2}\ \Leftrightarrow\dfrac{
1}{x+1}=\dfrac{1}{2}\
\Leftrightarrow x+1=2\
\Leftrightarrow x=1\
Vậy\text{                  }GTNN\text{                  }của\text{                  }phân\text{                  }thức\text{                  }là\text{                  }\dfrac{3}{4}\text{                  }khi\text{                  }x=1\)

\(\text{b)                      }\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\ =\dfrac{4x^2-4x-2x+1+1-1}{\left(2x-1\right)^2}\ =\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\ =1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\left(1\right)\ Đặt\text{                  }-\dfrac{1}{2x-1}=y\ \Rightarrow\left(1\right)=1+y+y^2\ =y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\ =\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\ =\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ Do\text{                  }\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\ \Rightarrow\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\ Dấu\text{                  }"="\text{                  }xảy\text{                  }ra\text{                  }khi:\
\left(y+\dfrac{1}{2}\right)^2=0\
\Leftrightarrow y+\dfrac{1}{2}=0\
\Leftrightarrow y=-\dfrac{1}{2}\
\Leftrightarrow-\dfrac{1}{2x-1}=-\dfrac{1}{2}\
\Leftrightarrow2x-1=2\
\Leftrightarrow2x=3\
\Leftrightarrow x=\dfrac{3}{2}\
Vậy\text{                     }GTNN\text{                     }của\text{                     }biểu\text{                     }thức\text{                     }là\text{                     }\dfrac{3}{4}\text{                     }khi\text{                     }x=\dfrac{3}{2}\)

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