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1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{x^4}{a}+\frac{y^4}{b}\right)(a+b)\geq (x^2+y^2)^2=1\)
\(\Leftrightarrow \frac{x^4}{a}+\frac{y^4}{b}\geq \frac{1}{a+b}\)
Dấu bằng xảy ra khi \(\frac{x^2}{a}=\frac{y^2}{b}\). Do đó \(\frac{x^2}{a}=\frac{y^2}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow \frac{x^{2006}}{a^{1003}}=\frac{y^{2006}}{b^{1003}}=\frac{1}{(a+b)^{1003}}\)
\(\Rightarrow \frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{y^{1003}}=\frac{2}{(a+b)^{1003}}\)
Do đó ta có đpcm.
Bài này phải quy đồng rồi áp dụng chớ chớ lỡ a+b=0 thì sao chị :3
a) \(9\left(2x-3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x-3\right)-2\left(x+1\right)\right]\left[3\left(2x-3\right)+2\left(x+1\right)\right]\)
\(=\left(6x-9-2x-2\right)\left(6x-9+2x+2\right)\)
\(=\left(4x-11\right)\left(8x-7\right)\)
b) \(\left(x^2+4y^2-20\right)-16\left(xy-4\right)^2\)
\(=\left[\left(x^2-4xy+4y^2\right)-4\right]\left[\left(x^2+4xy+4y^2\right)-36\right]\)
\(=\left[\left(x-2y\right)^2-4\right]\left[\left(x+2y\right)^2-36\right]\)
\(=\left(x-2y-2\right)\left(x-2y+2\right)\left(x+2y-6\right)\left(x+2y+6\right)\)
a. 9 ( 2x - 3 )2 - 4 ( x + 1 )2
= [ 3 ( 2x - 3 ) ]2 - [ 2 ( x + 1 ) ]2
= [ 3 ( 2x - 3 ) - 2 ( x + 1 ) ] [ 3 ( 2x - 3 ) + 2 ( x + 1 ) ]
= ( 6x - 9 - 2x - 2 ) ( 6x - 9 + 2x + 2 )
= ( 4x - 11 ) ( 8x - 7 )
b. ( x2 + 4y2 - 20 )2 - 16 ( xy - 4 )2
= ( x2 + 4y2 - 20 )2 - [ 4 ( xy - 4 ) ]2
= [ x2 + 4y2 - 20 - 4 ( xy - 4 ) ] [ x2 + 4y2 - 20 + 4 ( xy - 4 ) ]
= ( x2 + 4y2 - 20 - 4xy + 16 ) ( x2 + 4y2 - 20 + 4xy - 16 )
= ( x2 + 4y2 - 4xy - 4 ) ( x2 + 4y2 + 4xy - 36 )
= [ ( x - 2y )2 - 22 ] [ ( x + 2y )2 - 62 ]
= ( x - 2y - 2 ) ( x - 2y + 2 ) ( x + 2y - 6 ) ( x + 2y + 6 )
B=(2+1)(22+1)(24+1)(28+1)(216+1)−232
=1.(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(22-1)(22+1)(24+1)(28+1)(216+1)−232
=(24-1)(24+1)(28+1)(216+1)−232
=(28-1)(28+1)(216+1)−232
=(216-1)(216+1)−232
=232-1-232
=-1
A = ( 2 +1 )( 2^2 + 1 )...(2^16+1) - 2^32
A = ( 2 - 1) ( 2 + 1 )(2^2 + 1) .... (2^16 + 1) - 2^32
A = (2^2 - 1) (2^2 + 1) ...(2^16 + 1) - 2^32
A =( 2^ 4 - 1)( 2^4 + 1 )( 2^8 + 1) (2^16+1) -2^32
A = ( 2^8 - 1)( 2^ 8 + 1) ( 2^ 16 + 1)- 2^32
A = ( 2^16 - 1 )( 2^16 + 1) - 2^32
A = 2^32 - 1 - 2^32
A = - 1
a) 1272 + 146.127 + 732
= 1272 + 2.73.127 + 732
= (127 + 73)2 = 2002 = 40000
b) 98 . 28 - (184 - 1)(184 + 1)
= (9.2)8 - 188 + 1
= 188 - 188 + 1 = 1
c) \(\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{125^2+2.75.125+75^2}=\frac{560.1000}{\left(125+75\right)^2}=\frac{560000}{200^2}\)
\(=\frac{560000}{40000}=14\)
a) 1272 + 146.127 + 732
= 1272 + 2.73.127 + 732
= ( 127 + 73 )2
= 2002 = 40 000
b) 98.28 - ( 184 - 1 )( 184 + 1 )
= ( 9.2 )8 - [ ( 184 )2 - 12 ]
= 188 - 188 + 1
= 1
c) \(\frac{780^2-220^2}{125^2+150\cdot125+75^2}\)
\(=\frac{\left(780-220\right)\left(780+220\right)}{125^2+2\cdot75\cdot125+75^2}\)
\(=\frac{560\cdot1000}{\left(125+75\right)^2}\)
\(=\frac{560000}{200^2}\)
\(=\frac{560000}{40000}=14\)
Bài giải
\(a,\text{ }a^2+9-6a=a^2+2\cdot3a+3^2=\left(a-3\right)^2\)
\(b,\text{ }x^2-x+\frac{1}{4}=x^2-2\cdot\frac{1}{2}\cdot x+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)
\(c,\text{ }-x^2+4x-x=3x-x^2=\left(\sqrt{3x}\right)^2-x^2=\left(\sqrt{3x}-x\right)\left(\sqrt{3x}+x\right)\)( Đề nói vận dụng hằng đẳng thức để rút gọn nên mình đưa về hiệu hai ình phương nha ! )
a) = 1003 2 - 2.3.1003 + 32
= (1003 - 3)2 = 10002 = 1000000
b) = 9982 + 4. (998 + 1)
= 9982 + 2.2.998 + 22
= (998 + 2)2 = 10002 = 1000000
a) = 10032 - 2.3.1003 + 32
= (1003 - 3)2 = 10002 = 1000000
b) = 9982 + 4. (998 + 1)
= 9982 + 2.2.998 + 22
= (998 + 2)2 = 10002 = 1000000