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\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
Bài 1 :
x2-2x+2>0 với mọi x
=x2-2.x.1/4+1/16+31/16
=(x-1/4)2 + 31/16
Vì (x-1/4)2 \(\ge\) 0 nên (x-1/4)2 + 31/16 \(\ge\) 0 với mọi x (đfcm)
bài 1) a) \(A=\left(2x-5\right)^2-4\left(2x-5\right)+5=4x^2-20x+25-8x+20+5\)
\(A=4x^2-28x+49+1=\left(2x-7\right)^2+1\ge1\forall m\)(đpcm)
b) \(A=\left(2x-7\right)^2+1\ge1\) \(\Rightarrow minA=1\Leftrightarrow\left(2x-7\right)^2=0\Leftrightarrow2x-7=0\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\)
Bài 1:
a)\(A=\left(2x-5\right)^2-4\left(2x-5\right)+5\)
\(=\left(2x-5\right)^2-4\left(2x-5\right)+4+1\)
\(=\left(2x-5-2\right)^2+1\)
\(=\left(2x-7\right)^2+1\ge1\)
b)Xảy ra khi \(\left(2x-7\right)^2=0\)
\(\Rightarrow2x-7=0\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
Bài 2:
a)\(B=-\left(3x+7\right)^2+2\left(3x+7\right)-17\)
\(=-\left(3x+7\right)^2+2\left(3x+7\right)-1-16\)
\(=-\left[\left(3x+7\right)^2-2\left(3x+7\right)+1\right]-16\)
\(=-\left(3x+7-1\right)^2-16\)
\(=-\left(3x+6\right)^2-16\le-16\)
b)Xảy ra khi \(-\left(3x+6\right)^2=0\)
\(\Rightarrow3x+6=0\Rightarrow3x=-6\Rightarrow x=-2\)