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=a, (x-3)(x+3)-(x-7)(x+7)= x2 - 9 - x2 + 7
= -2
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)= (4x-5)2 - 2(4x+5)(3x-2) + (3x-2)2
= ( 4x - 5 - 3x + 2 )2
= ( x - 3 )2
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2= 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= (3x-y)2+ 2(3x-y)(3x+y)+ (3x+y)2
= ( 3x - y + 3x + y )2
= ( 6x )2
= 36x2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
1, rút gọn
a, (x-3)(x+3)-(x-7)(x+7)
= x^2 - 9 - (x^2 - 49)
= x^2 - 9 - x^2 + 49
= 40
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)
= 16x^2 - 40x + 25 + 9x^2 - 12x + 4 - 2(12x^2 - 8x + 15x - 10)
= 25x^2 - 52x + 29 - 24x^2 + 16x - 30x + 20
= x^2 - 66x + 49
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= 2(9x^2 - y^2) + 9x^2 - 6xy + y^2 + 9x^2 + 6xy + y^2
= 18x^2 - 2y^2 + 18x^2 + 2y^2
= 36x^2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
= dài vl
Ta có (21 -1)(21 + 1) = 22 - 1
(22 - 1)(22 + 1) = 24 - 1
tương tự như vậy ta sẽ có (2 -1)A = 232 - 1
vậy A < 232
Ta có: \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}\)
\(\Leftrightarrow A< B\)
A= \(\frac{3\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{\left(2^2-1\right)}=2^{32-1}\)
mà B= \(2^{32}\)
=> A<B
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
Bài 1:
a) \(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)
b) Sửa đề \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)-3^{64}\)
\(=3^{64}-1-3^{64}\)
\(=-1\)
Bài 2:
Ta có:
\(A=2009.2009\)
\(A=2009\left(2008+1\right)\)
\(A=2009.2008+2009\)
Ta lại có:
\(B=2008.2010\)
\(B=2008\left(2009+1\right)\)
\(B=2008.2009+2008\)
Vì 2008.2009 = 2009.2008
2009 > 2008
=> 2008.2009 + 2009 > 2009.2008 + 2008
=> A > B
1,a,(2-1)(2+1)(22+1)(24+1)(28+1)
=(22-1)(22+1)(24+1)(28+1)
=(24-1) (24+1)(28+1)
=(28 -1)(28+1)=216-1
2,
A=2009.2009=20092
B=2008.2010=(2009-1)(2009+1)=20092-1
Do20092>20092-1\(\Rightarrow A>B\)
A = (2 - 1)(2 + 1)(2^2 + 1 )(2^4 + 1 ) (2^8 + 1)(2^16 + 1) ( nhân vói 2 - 1 = 1 Gía không thay dổi)
A = ( 2 ^2 - 1 )(2^2 + 1 )(2^4 + 1 )(2^8 + 1 )(2^16 + 1 )
A = ( 2^4 - 1 )(2^4 + 1)(2^8 + 1)(2^16 + 1)
A = (2^8 - 1)(2^8 + 1)(2^16 + 1)
A = (2^16 - 1)(2^16 + 1 )
A = 2^32 - 1 <2^32 = B
VẬy A < B
\(a)\) Ta có :
\(A=2008.2010=\left(2009+1\right)\left(2009-1\right)=2009^2-1< 2009^2=B\)
Vậy \(A< B\)
\(b)\)\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\) (bn xem lại đề xem có nhầm j ko, nếu đề đúng thì mk sr)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(A=2^{32}-1< 2^{32}=B\)
Vậy \(A< B\)
Chúc bạn học tốt ~