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Đặt A = ( 2 + 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 )
=> ( 2 - 1 )A = ( 2 - 1 )( 2 + 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 )
=> A = ( 22 - 1 )( 22 + 1 )( 24 + 1 )...( 2256 + 1 )
= ( 24 - 1 )( 24 + 1 )...( 2256 + 1 )
= ( 2256 - 1 )( 2256 + 1 )
= 2512 - 1
Câu 1
5x2 + 10y2 - 6xy - 4x - 2y + 3
= ( x2 - 6xy + 9y2 ) + ( 4x2 - 4x + 1 ) + ( y2 - 2y + 1 ) + 1
= ( x - 3y )2 + ( 2x - 1 )2 + ( y - 1 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
Câu 2
a) A = 2011.2013 = ( 2012 - 1 )( 2012 + 1 ) = 20122 - 1 < 20122
=> A < B
B = 3128 - 1
= ( 364 - 1 )( 364 + 1 )
= ( 332 - 1 )( 332 + 1 )( 364 + 1 )
= ( 316 - 1 )( 316 + 1 )( 332 + 1 )( 364 + 1 )
= ( 34 - 1 )( 34 + 1 )( 316 + 1 )( 332 + 1 )( 364 + 1 )
= ( 32 - 1 )( 32 + 1 )( 34 + 1 )( 316 + 1 )( 332 + 1 )( 364 + 1 )
= ( 3 - 1 )( 3 + 1 )( 32 + 1 )( 34 + 1 )( 316 + 1 )( 332 + 1 )( 364 + 1 )
= 8( 32 + 1 )( 34 + 1 )( 316 + 1 )( 332 + 1 )( 364 + 1 ) > 4( 32 + 1 )( 34 + 1 )( 316 + 1 )( 332 + 1 )( 364 + 1 )
=> B > A
2.Tim x
a,(2x+1)2-4(x+2)2=9
<=> (4x2+4x+1)-4(x2+4x+4)=9
<=> -12x-15=9
<=> -12x=24
<=> x=-2
\(1a,\)\(\left(x^2-0,1\right)=\left(x-\sqrt{0,1}\right)\left(x+\sqrt{0,1}\right)\)
\(1b,\)\(\left(2a^2+b^2\right)^2=\left(2a^2\right)^2+2.2a^2.b^2+\left(b^2\right)^2=4a^4+4a^2b^2+b^4\)
\(1c,\)\(\left(a^2+5\right)\left(5-a^2\right)=\left(5+a^2\right)\left(5-a^2\right)=25-x^4\)
mk giải từng nha == tại vì mk sợ nhiều qus bị troll
\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(27x^3+18x^2+12x-18x^2-12x-8-3x\left(9x^2-3x+1\right)+\left(9x^2-3x+1\right)=x-4\)
\(27x^3-8-3\left(9x^2-3x+1\right)+9x^2-3x+1=x-4\)
\(27x^3-7-3x\left(9x^2-3x+1\right)+9x^2-3x=x-4\)
\(27x^3-7-27x^3+9x^2-3x+9x^2-3x=x-4\)
\(-7+18x^2-6x=x-4\)
\(3-18x^2+7x=0\)
\(x=\frac{-7+\sqrt{265}}{-36};\frac{-7-\sqrt{265}}{-36}\)
\(9\left(2x+1\right)=4\left(x-5\right)^2\)
\(18x+9=4x^2-40x+100\)
\(18x+9-4x^2+40x-100=0\)
\(58x-91-4x^2=0\)
\(x=\frac{29-3\sqrt{53}}{4};\frac{29+3\sqrt{53}}{4}\)
Câu hỏi của Trịnh Minh Châu - Toán lớp 8 - Học toán với OnlineMath
a) 7x+7y=7(x+y)
b) 2x2y-6xy2=2xy(x-3y)
c)3x(x-1)+7x2(x-1)=x(x-1)(3+7x)
d)3x(x-4)+5x2(4-x)=(x-4)(3x-5x2)
=x(x-4)(3-5x)
e)6x4-9x3=3x3(2x-3)
f)5y8-15y6=5y6(y2-3)
\(=1.\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)=\left(2-1\right)\left(2+1\right)...\left(2^{256}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{256}+1\right)=\left(2^4-1\right)...\left(2^{256}+1\right)=....=2^{512}-1\)
A= (2-1) (2+1) (22+1) ........ + 1
= (22-1)(22+1) ......... (2256+1)
= (24-1) (24+ 1) ......... (2256+1)
................
= [(2256)2 –1] + 1
= 2512