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a)
\(x^2+xy+y^2+1=\left(x^2+2x\times\frac{y}{2}+\left(\frac{y}{2}\right)^2\right)+\frac{3y^2}{4}+1\)
\(=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\ge0+0+1=1\)
mà\(1>0\Rightarrow x^2+xy+y^2+1>0\)với mọi \(x\)và\(y\)
b)
\(x^2+5y^2+2x-4xy-10y+14\)
\(=\left[x^2+2x\left(1-2y\right)+\left(1-2y\right)^2\right]+y^2-6y+13\)
\(=\left(x+1-2y\right)^2+\left(y^2-2y\times3+9\right)+4\)
\(=\left(x+1-2y\right)^2+\left(y-3\right)^2+4\)
Ta có:\(\left(x+1-2y\right)^2\ge0\)với mọi \(x;y\in R\)
và\(\left(y-3\right)^2\ge0\)với mọi \(x;y\in R\)
\(\Rightarrow\left(x+1-2y\right)^2+\left(y-3\right)^2+4\ge4\)với mọi \(x;y\in R\)
\(\Rightarrow x^2+5y^2+2x-4xy-10y+14>0\)
c)
\(5x^2+10y^2-6xy-4x-2y+3=x^2+4x^2+y^2+9y^2-6xy-4x-2y+3\)
\(=\left[\left(2x\right)^2-2\times2x+1\right]+\left(y^2-2y+1\right)+\left[\left(3y\right)^2-2\times3y+x^2\right]+1\)
\(=\left(2x+1\right)^2+\left(y-1\right)^2+\left(3y-x\right)^2+1\)
Ta có \(\left(2x+1\right)^2\ge0\)với mọi \(x\)
\(\left(y-1\right)^2\ge\)với mọi \(y\)
\(\left(3y-x\right)^2\ge0\)với mọi \(x;y\)
và \(1>0\)
\(\Rightarrow5x^2+10y^2-6xy-4x-2y+3>0\)
a. \(x^2+xy+y^2+1=\left(x^2+xy+\frac{1}{4}y^2\right)+\frac{3}{4}y^2+1=\left(x+\frac{1}{4}y\right)^2+\frac{3}{4}y^2+1>0\forall x;y\)(đpcm)
b. \(x^2+5y^2+2x-4xy-10y+14\)
\(=\left[\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1\right]+\left(y^2-6y+9\right)+4\)
\(=\left[\left(x-2y\right)^2-2\left(x-2y\right)+1\right]+\left(y^2-6y+9\right)+4\)
\(=\left(x-2y-1\right)^2+\left(y-3\right)^2+4>0\forall x;y\)(đpcm)
c. tương tự ý b
13.
M \(=\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)\)\(+16\)
\(=\)\(\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)\) \(+16\)
\(=\left(x^2+10x+20\right)^2-16+16\)
\(=\left(x^2+10x+20\right)^2\) là một số chính phương
Nhiều quá, nhìn đã thấy ớn lạnh :(
Bạn nên chia nhỏ ra , post 1 hoặc 2 bài 1 lần thôi, đăng 1 lần 1 nùi thế này không ai dám làm đâu, bội thực chữ viết.
a/ \(x^2+xy+y^2+1=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}+1=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1>0\)
b/ \(x^2+5y^2+2x-4xy-10y+14\)
\(=\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1+\left(y^2-6y+9\right)+4\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)
C1. ( 2x + 3y )2 + 2( 2x + 3y ) + 1 = [ ( 2x + 3y ) + 1 ]2
C2. ( x + 2 )2 = ( 2x - 1 )2
<=> ( x + 2 )2 - ( 2x - 1 )2 = 0
<=> [ x + 2 + ( 2x - 1 ) ][ x + 2 - ( 2x - 1 ) ] = 0
<=> [ 3x + 1 ][ 3 - x ] = 0
<=> \(\orbr{\begin{cases}3x+1=0\\3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=3\end{cases}}\)
b) ( x + 2 )2 - x + 4 = 0
<=> x2 + 4x + 4 - x + 4 = 0
<=> x2 - 3x + 8 = 0
Mà ta có x2 - 3x + 8 = x2 - 3x + 9/4 + 23/4 = ( x - 3/2 )2 + 23/4 ≥ 23/4 > 0 với mọi x
=> Phương trình vô nghiệm
C3. a) A = x2 - 2x + 5 = x2 - 2x + 4 + 1 = ( x - 2 )2 + 1
\(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2+1\ge1\)
Dấu " = " xảy ra <=> x - 2 = 0 => x = 2
Vậy AMin = 1 , đạt được khi x = 2
b)B = x2 - x + 1 = x2 - x + 1/4 + 3/4 = ( x - 1/2 )2 + 3/4
\(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu " = " xảy ra <=> x - 1/2 = 0 => x = 1/2
Vậy BMin = 3/4, đạt được khi x = 1/2
c) C = ( x - 1 )( x + 2 )( x + 3 )( x + 6 )
C = [ ( x - 1 )( x + 6 )][ ( x + 2 )( x + 3 ]
C = [ x2 + 5x - 6 ][ x2 + 5x + 6 ]
C = ( x2 + 5x )2 - 36
\(\left(x^2+5x\right)^2\ge0\forall x\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
Dấu " = " xảy ra <=> x2 + 5x = 0
<=> x( x + 5 ) = 0
<=> x = 0 hoặc x + 5 = 0
<=> x = 0 hoặc x = -5
Vậy CMin = -36, đạt được khi x = 0 hoặc x = -5
d) D = x2 + 5y2 - 2xy + 4y + 3
= ( x2 - 2xy + y2 ) + ( 4y2 + 4y + 1 ) + 2
= ( x - y )2 + ( 2y + 1 )2 + 2
\(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(2y+1\right)^2\ge0\end{cases}}\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2\ge0\forall x,y\)
=> \(\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-y=0\\2y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x-y=0\\y=-\frac{1}{2}\end{cases}\Rightarrow}x=y=-\frac{1}{2}\)
Vậy DMin = 2 , đạt được khi x = y = -1/2
C4. a) ( Cái này tìm được Min k tìm được Max )
A = x2 - 4x - 2 = x2 - 4x + 4 - 6 = ( x - 2 )2 - 6
\(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2-6\ge-6\)
Dấu " = " xảy ra <=> x - 2 = 0 => x = 2
Vậy AMin = -6 , đạt được khi x = 2
b) B = -2x2 - 3x + 5 = -2( x2 + 3/2x + 9/16 ) + 49/8 = -2( x + 3/4 )2 + 49/8
\(-2\left(x+\frac{3}{4}\right)^2\le0\Rightarrow-2\left(x+\frac{3}{4}\right)+\frac{49}{8}\le\frac{49}{8}\)
Dấu " = " xảy ra <=> x + 3/4 = 0 => x = -3/4
Vậy BMax = 49/8 , đạt được khi x = -3/4
c) C = ( 2 - x )( x + 4 ) = -x2 - 2x + 8 = -( x2 + 2x + 1 ) + 9 = -( x + 1 )2 + 9
\(-\left(x+1\right)^2\le0\Rightarrow-\left(x+1\right)^2+9\le9\)
Dấu " = " xảy ra <=> x + 1 = 0 => x = -1
Vậy CMax = 9 , đạt được khi x = -1
d) D = -8x2 + 4xy - y2 + 3 ( Cái này mình đang tính ạ )
C5. a) A = 25x2 - 20x + 7
A = 25x2 - 20x + 4 + 3
A = ( 5x2 - 2 )2 + 3 ≥ 3 > 0 với mọi x ( đpcm )
b) B = 9x2 - 6xy + 2y2 + 1
B = ( 9x2 - 6xy + y2 ) + y2 + 1
B = ( 3x - y )2 + y2 + 1 ≥ 1 > 0 với mọi x, y ( đpcm )
c) C = x2 - 2x + y2 + 4y + 6
C = ( x2 - 2x + 1 ) + ( y2 + 4y + 4 ) + 1
C = ( x - 1 )2 + ( y + 2 )2 + 1 ≥ 1 > 0 với mọi x,y ( đpcm )
d) D = x2 - 2x + 2
D = x2 - 2x + 1 + 1
D = ( x - 1 )2 + 1 ≥ 1 > 0 với mọi x ( đpcm )
\(A=16x^2+8x+3\\ A=16x^2+8x+1+2\\ A=\left(16x^2+8x+1\right)+2\\ A=\left(4x+1\right)^2+2\\ Do\left(4x+1\right)^2\ge0\forall x\\ \Rightarrow A=\left(4x+1\right)^2+2\ge2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(4x+1\right)^2=0\\ \Leftrightarrow4x+1=0\\ \Leftrightarrow4x=-1\\ \Leftrightarrow x=-\dfrac{1}{4}\\ \text{Vậy }A_{\left(Min\right)}=2\text{ khi }x=-\dfrac{1}{4}\\ \)
\(B=y^2-5y+8\\ B=y^2-5y+\dfrac{25}{4}+\dfrac{7}{4}\\ B=\left(y^2-5y+\dfrac{25}{4}\right)+\dfrac{7}{4}\\ B=\left[y^2-2\cdot y\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{7}{4}\\ B=\text{ }\left(y-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\\ Do\text{ }\left(y-\dfrac{5}{2}\right)^2\ge0\forall x\\ \Rightarrow B=\left(y-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(y-\dfrac{5}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{5}{2}=0\\ \Leftrightarrow y=\dfrac{5}{2}\\ \text{Vậy }B_{\left(Min\right)}=\dfrac{7}{4}\text{ }khi\text{ }y=\dfrac{5}{2}\)
\(C=2x^2-2x+2\\ C=2x^2-2x+\dfrac{1}{2}+\dfrac{3}{2}\\ C=\left(2x^2-2x+\dfrac{1}{2}\right)+\dfrac{3}{2}\\ C=2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{3}{2}\\ C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{2}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{1}{2}=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{3}{2}\text{ }khi\text{ }x=\dfrac{1}{2}\)
\(D=9x^2-6x+25y^2+10y+4\\ D=9x^2-6x+25y^2+10y+1+1+2\\ D=\left(9x^2-6x+1\right)+\left(25y^2+10y+1\right)+2\\ D=\left[\left(3x\right)^2-2\cdot3x\cdot1+1^2\right]+\left[\left(5y\right)^2+2\cdot5y\cdot1+1^2\right]+2\\ D=\left(3x-1\right)^2+\left(5y+1\right)^2+2\\ Do\text{ }\left(3x-1\right)^2\ge0\forall x\\ \left(5y+1\right)^2\ge0\forall y\\ \Rightarrow\left(3x-1\right)^2+\left(5y+1\right)^2\ge0\forall x;y\\ \Rightarrow D=\left(3x-1\right)^2+\left(5y+1\right)^2+2\ge2\forall x;y\\ \text{Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(3x-1\right)^2=0\\\left(5y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\5y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=1\\5y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{5}\end{matrix}\right.\\ \text{Vậy }D_{\left(Min\right)}=2\text{ khi }x=\dfrac{1}{3};y=-\dfrac{1}{5}\)
Câu 2
\(M=x^2+6x+1\\ M=x^2+6x+9-8\\ M=\left(x^2+6x+9\right)-8\\ M=\left(x+3\right)^2-8\\ Do\text{ }\left(x+3\right)^2\ge0\forall x\\ M=\left(x+3\right)^2-8\ge-8\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\\ \text{Vậy }M_{\left(Min\right)}=-8\text{ khi }x=-3\)
\(N=10y-5y^2-3\\ N=10y-5y^2-5+2\\ N=-\left(5y^2-10y+5\right)+2\\ N=-5\left(y^2-2y+1\right)+2\\ N=-5\left(y-1\right)^2+2\\ Do\left(y-1\right)^2\ge0\forall x\\ \Rightarrow-\left(y-1\right)^2\le0\forall x\\ \Rightarrow-5\left(y-1\right)^2\le0\forall x\\ \Rightarrow N=-5\left(y-1\right)^2+2\le2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(y-1\right)^2=0\\ \Leftrightarrow y-1=0\\ \Leftrightarrow y=1\\ \text{Vậy }N_{\left(Max\right)}=2\text{ khi }y=1\)
a: \(VT=x^2+2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1\)
\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1>0\forall x,y\)
c: \(VT=x^2-6xy+9y^2+4x^2-4x+1+y^2-2y+1+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\forall x,y\)
3)
e)
b) Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0