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\(A=4x^2+4x+11\)
\(=\left(4x^2+4x+1\right)+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Min A = 10 khi: 2x + 1 = 0
<=> x = -1/2
a/ \(4x^2+4x+11\)
\(=\left(2x^2\right)+2\cdot2x+1-1+11\)
\(=\left(2x+1\right)^2-1+11\)
\(=\left(2x+1\right)^2+10\)
Có : \(\left(2x+1\right)^2\ge0\)
\(\Rightarrow\left(2x+1\right)^2+10\ge10\)
\(\Rightarrow GTNN\left(4x^2+4x+11\right)=10\)
Với \(\left(2x+1\right)^2=0;x=-\frac{1}{2}\)
\(a,A=4x^2+4x+11\)
\(A=(2x+1)^2+10\)
Do \((2x+1)^2\ge0\Rightarrow(2x+1)^2+10\ge10\forall x\)
\(\Rightarrow Min_a=10\Rightarrow2x+1=0\Rightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)
Vậy giá trị nhỏ nhất của A là 10 khi x = -1/2
1)
\(a,\) \(A=4x^2+4x+11\)
\(=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy : min \(A=10\Leftrightarrow x=-\frac{1}{2}\)
b) \(C=x^2-2x+y^2-4y+7\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow x=1,y=2\)
Vậy : \(minC=2\Leftrightarrow x=1,y=2\)
2,
a) \(A=5-8x-x^2\)
\(=-\left(x^2+8x+16\right)+21=-\left(x+4\right)^2+21\le21\)
Dấu "=" xảy ra \(\Leftrightarrow x=-4\)
b) \(B=5-x^2+2x-4y^2-4y\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)
Dấu "=" xảy ra \(\Leftrightarrow x=1,y=-\frac{1}{2}\)
a) Ta có: \(A=4x^2+4x+11\)
\(\Rightarrow A=4x^2+2x+2x+11\)
\(\Rightarrow A=2x.\left(2x+1\right)+\left(2x+1\right)+10\)
\(\Rightarrow A=\left(2x+1\right).\left(2x+1\right)+10\)
\(\Rightarrow A=\left(2x+1\right)^2+10\)
Ta lại có: \(\left(2x+1\right)^2\ge0\forall x\inℝ\)
\(\Rightarrow A\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vậy \(A_{min}=10\Leftrightarrow x=\frac{-1}{2}\)
Bài làm:
a) Ta có: \(A=4x^2+4x+11=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(2x+1\right)^2=0\Rightarrow x=-\frac{1}{2}\)
Vậy \(Min_A=10\Leftrightarrow x=-\frac{1}{2}\)
b) \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(B=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(B=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(B=\left(x^2+5x\right)^2-36\ge-36\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x^2+5x\right)^2=0\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy \(Min_B=-36\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
c) Ta có: \(C=x^2-2x+y^2-4y+7\)
\(C=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)
\(C=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy \(Min_C=2\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)
a) A = 4x2 + 4x + 11
A = 4( x2 + x + 1/4 ) + 10
A = 4( x + 1/2 )2 + 10
\(4\left(x+\frac{1}{2}\right)^2\ge0\forall x\Rightarrow4\left(x+\frac{1}{2}^2\right)+10\ge0\)
Dấu " = " xảy ra <=> x + 1/2 = 0 => x = -1/2
Vậy AMin = 10 , đạt được khi x = -1/2
b) B = ( x - 1 )( x + 2 )( x + 3 )( x + 6 )
B = [( x - 1 )( x + 6 )][( x + 2 )( x + 3 )]
B = ( x2 + 5x - 6 )( x2 + 5x + 6 )
Đặt a = x2 + 5x
=> B = ( a - 6 )( a + 6 ) = a2 - 36
\(a^2\ge0\forall a\Rightarrow a^2-36\ge-36\)
Dấu " = " xảy ra <=> a2 = 0 => a = 0
<=> x2 + 5x = 0
<=> x( x + 5 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy BMin = -36 , đạt được khi x = 0 hoặc x = -5
c) C = x2 - 2x + y2 - 4y + 7
C = ( x2 - 2x + 1 ) + ( y2 - 4y + 4 ) + 2
C = ( x - 1 )2 + ( y - 2 )2 + 2
\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy CMin = 2 , đạt được khi x = 1, y = 2
( x - 1) ( x + 6 ) ( x + 2 ) ( x + 3 )
<=> ( x2 + 6x - x - 6 ) ( x2 + 3x + 2x + 6)
<=> ( x2 - 5x )2 lun nhỏ hơn 0
Nên dấu " =" xảy ra khi ( x2- 5x)2 = 0
x2 - 5x= 0 <=> x ( x - 5) = 0 <=> x=0 hoặc 5
^^ Học tốt nha!!!!
\(4.\)
\(a.A=5-8x-x^2\)
\(=-\left(16+8x+x^2\right)+21\)
\(=-\left(4+x\right)^2+21\le21\)
\(A_{max}=21\)
Dấu '='xảy ra khi \(x=-4\)
\(b.B=5-x^2+2x-4y^2-4y\)
\(=-\left(1-2x+x^2\right)-\left(4+4y+4y^2\right)+10\)
\(=-\left(1-x\right)^2-\left(2+2y\right)^2+10\le10\)
\(B_{max}=10\)
Dấu '=' xảy ra khi \(x=1;y=-1\)
\(5.\)
\(a.\) Ta có:\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a-b=0\Leftrightarrow a=b\left(1\right)\)
hay\(b-c=0\Leftrightarrow b=c\left(2\right)\)
hay\(c-a=0\Leftrightarrow c=a\left(3\right)\)
Từ \(\left(1\right),\left(2\right)\)và\(\left(3\right)\)suy ra:\(a=b=c\left(đpcm\right)\)
\(b.a^2-2a+b^2+4b+4c^2-4c+6=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)
\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)
hay\(b+2=0\Leftrightarrow b=-2\)
hay\(2c-2=0\Leftrightarrow c=1\)
V...
^^
Bài 1:
a, \(A=4x^2+4x+1\)
\(A=4x^2+2x+2x+1\)
\(A=2x.\left(2x+1\right)+\left(2x+1\right)\)
\(A=\left(2x+1\right)^2\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(2x+1\right)^2\ge0\)
Hay \(A\ge0\) với mọi giá trị của \(x\in R\).
Để \(A=0\)thì \(\left(2x+1\right)^2=0\Rightarrow2x=-1\Rightarrow x=\dfrac{-1}{2}\)
Vậy.....
b, \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(B=\left[\left(x-1\right).\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(B=\left(x^2+6x-x+6\right).\left(x^2+3x+2x+6\right)\)
\(B=\left(x^2+5x+6\right)\left(x^2+5x+6\right)\)
\(B=\left(x^2+5x+6\right)^2\)
\(B=\left(x^2+2,5x+2,5x+6,25-0,25\right)^2\)
\(B=\left[\left(x+2,5\right)^2-0,25\right]^2\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x+2,5\right)^2\ge0\Rightarrow\left(x+2,5\right)^2-0,25\ge-0,25\)
\(\Rightarrow\left[\left(x+2,5\right)^2-0,25\right]^2\ge0,0625\)
Hay \(B\ge0,0625\) với mọi giá trị của \(x\in R\).
Để \(B=0,0625\) thì \(\left[\left(x+2,5\right)^2-0,25\right]^2=0,0625\)
\(\Rightarrow\left(x+2,5\right)^2-0,25=0,25\)
\(\Rightarrow x+2,5=0\Rightarrow x=-2,5\)
Vậy.......
Câu c làm tương tự!! Chúc bạn học tốt!!!
\(A=4x^2+4x+1=\left(2x+1\right)^2\ge0\)
Vậy GTNN của A là 0 khi \(\left(2x+1\right)^2=0\Rightarrow2x+1=0\Rightarrow x=\dfrac{-1}{2}\)
\(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\) \(=\left(x^2+5x\right)^2-36\ge-36\)
Vậy GTNN của B là -36 khi \(\left(x^2+5x\right)^2=0\Rightarrow x\left(x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\) \(C=x^2-2x+y^2-4y+7=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+3=\left(x-1\right)^2+\left(y-2\right)^2+3\ge3\)
Vậy GTNN của C là 3 khi \(\left[{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)