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a) A= \(\left(x^2-2xy+y^2\right)+\left(x^2+10x+25\right)+x^2+1\)1
=\(\left(x-y\right)^2+\left(x+5\right)^2+x^2+1\ge1\)
\(\Rightarrow\)A dương với mọi x,y
Z để 2n2 + 7n -2 chia hết cho 2n-1
1/ Sửa đề a+b=1
\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
Thay a+b=1 vào M ta được:
\(M=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2/ Đặt \(A=\frac{2n^2+7n-2}{2n-1}=\frac{\left(2n^2-n\right)+\left(8n-4\right)+2}{2n-1}=\frac{n\left(2n-1\right)+4\left(2n-1\right)+2}{2n-1}=n+4+\frac{2}{2n-1}\)
Để \(A\in Z\Leftrightarrow2n-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Ta có bảng:
| 2n-1 | 1 | -1 | 2 | -2 |
| n | 1 | 0 | 3/2 (loại) | -1/2 (loại) |
Vậy n={1;0}
1) \(A=x\left(x-6\right)+10=x^2-6x+10=x^2-6x+9+1=\left(x-3\right)^2+1\ge1>0\)
Dấu "=" xảy ra khi: \(x=3\)
\(B=x^2-2x+9y^2-6y+3\)
\(B=\left(x^2-2x+1\right)+\left(9y^2-6y+1\right)+1\)
\(B=\left(x-1\right)^2+\left(3y-1\right)^2+1\ge1>0\)
Dấu "=" xảy ra khi: \(x=y=1\)
2) \(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi: \(x=2\)
\(B=4x^2+4x+11=4x^2+4x+1+10=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi: \(x=-\dfrac{1}{2}\)
\(C\) mk nghĩ đề sai
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(C=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)\)
\(C=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)\)
\(C=\left(x^2+5x+5\right)^2-1\)
\(C=\left(x^2+5x+\dfrac{25}{4}-\dfrac{5}{4}\right)^2-1\)
\(C=\left[\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}\right]^2-1\ge\dfrac{9}{16}\)
Dấu "=" xảy ra khi: \(x=-\dfrac{5}{2}\)
\(D=4x-x^2+1=-\left(x^2-4x-1\right)=-\left(x^2-4x+4-5\right)=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)
Dấu "=" xảy ra khi: \(x=2\)
\(E=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)=-\left(x+4\right)^2+21\le21\)
Dấu "=" xảy ra khi: \(x=-4\)
\(D=x^2-4x-3\)
\(D=x^2-4x+4-7\)
\(D=\left(x-2\right)^2-7\ge-7\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=2\)
\(E=x^2-6x+1\)
\(E=x^2-6x+9-8\)
\(E=\left(x-3\right)^2-8\ge-8\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
\(F=x^2+x+1\)
\(F=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(F=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{-1}{2}\)
\(G=x^2+x\)
\(G=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)
\(G=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{-1}{2}\)
\(H=2x^2-4x+2018\)
\(H=2\left(x^2-2x+1009\right)\)
\(H=2\left(x^2-2x+1+1008\right)\)
\(H=2\left[\left(x-1\right)^2+1008\right]\)
\(H=2\left(x-1\right)^2+2016\ge2016\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
\(I=2x^2+y^2+2x+2xy+2019\)
\(I=\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+2018\)
\(I=\left(x+y\right)^2+\left(x+1\right)^2+2018\ge2018\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
a) x2 + x + 1 = ( x2 + x + 1/4 ) + 3/4 = ( x + 1/2 )2 + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )
b) 4x2 - 2x + 1 = 4( x2 - 1/2x + 1/16 ) + 3/4 = 4( x - 1/4 )2 + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )
c) x4 - 3x2 + 9 (*)
Đặt t = x2
(*) <=> t2 - 3t + 9 = ( t2 - 3t + 9/4 ) + 27/4 = ( t - 3/2 )2 + 27/4 = ( x2 - 3/2 )2 + 27/4 ≥ 27/4 > 0 ∀ x ( đpcm )
d) x2 + y2 - 2x - 4y + 6 = ( x2 - 2x + 1 ) + ( y2 - 4y + 4 ) + 1 = ( x - 1 )2 + ( y - 2 )2 + 1 ≥ 1 > 0 ∀ x, y ( đpcm )
e) x2 + y2 - 2x - 2y + 2xy + 2 = ( x2 + 2xy + y2 - 2x - 2y + 1 ) + 1
= [ ( x2 + 2xy + y2 ) - ( 2x + 2y ) + 1 ] + 1
= [ ( x + y )2 - 2( x + y ) + 12 ] + 1
= ( x + y - 1 )2 + 1 ≥ 1 > 0 ∀ x, y ( đpcm )
a) \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)
b) \(4x^2-2x+1=4\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{3}{4}=4\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)
c) \(x^4-3x^2+9=\left(x^4-3x^2+\frac{9}{4}\right)+\frac{27}{4}=\left(x^2-\frac{3}{2}\right)^2+\frac{27}{4}>0\left(\forall x\right)\)
d) \(x^2+y^2-2x-4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\left(\forall x,y\right)\)
e) \(x^2+y^2-2x-2y+2xy+2\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1+1\)
\(=\left(x+y-1\right)^2+1>0\left(\forall x,y\right)\)
a) \(x^2 +x +1 = x^2 +x +1/4 +3/4 = (x+1/2)^2 +3/4\)
các câu khác dùng phương pháp tương tự
a) x^2 + x +1 = x^2 + x + 1/4 + 3/4 = ( x+ 1/2)^2 + 3/4
Vì (x+1/2)^2 >= 0 => (x+1/2)^2 + 3/4>=3/4 > 0
b) 4x^2 - 2x + 1 = (2x)^2 - 2x + 1/4 + 3/4 = (2x +1/2)^2 + 3/4
Vì (2x +1/2)^2 >=0 => (2x +1/2)^2 + 3/4 >= 3/4 > 0
c) x^4 -3x^2 + 9 = x^4 - 3x^2 + 9/4 + 25/4 = ( x^2+ 3/2)^2 + 9/4
Vì ( x^2+ 3/2)^2 >= 0 => ( x^2+ 3/2)^2 + 9/4 >=9/4 >0
d) x^2 + y^2 -2x-2y + 2xy +1
= ( x^2 + 2xy + y^2) - 2( x+y) +1
= ( x+y)^2 -2(x+y) +1
= (x +y +1)^2 >=0
g) x^2+y^2+2(x-2y)+6
= (x^2 + 2x +1) + (y^2 -4y+4) +1
= ( x+1)^2 + (y-2)^2 +1
Vì (x+1)^2; (y-2)^2 >= 0 => ( x+1)^2 + (y-2)^2 +1>=1>0
A=x2-6x+10
\(A=\left(x-3\right)^2+1>1\)
\(\Rightarrow A\) luôn dương
A = x2 - 6x + 10
= ( x2 - 6x + 9 ) + 1
= ( x - 3 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
B = x2 + x + 5
= ( x2 + x + 1/4 ) + 19/4
= ( x + 1/2 )2 + 19/4 ≥ 19/4 > 0 ∀ x ( đpcm )
C = 4x2 + 4x + 2
= 4( x2 + x + 1/4 ) + 1
= 4( x + 1/2 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
D = ( x - 3 )( x - 5 ) + 4
= x2 - 8x + 15 + 4
= ( x2 - 8x + 16 ) + 3
= ( x - 4 )2 + 3 ≥ 3 > 0 ∀ x ( đpcm )
E = x2 - 2xy + 1 + y2
= ( x2 - 2xy + y2 ) + 1
= ( x - y )2 + 1 ≥ 1 > 0 ∀ x, y ( đpcm )