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a) \(=\left[\left(6x+1\right)+\left(6x-1\right)\right]^2\)
\(=\left(12x\right)^2\)
\(=144x^2\)
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
b)$3(2^2-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3(2^2-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3(2^2-1)\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3(2^2-1)\cdot\dfrac{2^{32}-1}{2^4-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2+1}$
$=\dfrac{3(2^{32}-1)}{5}$
a ) ( 6x + 1 )2 + ( 6x - 1 )2 - 2 . ( 6x + 1 )( 6x - 1 )
= ( 6x + 1 )2 - 2 . ( 6x + 1 )( 6x - 1 ) + ( 6x - 1 )2
= ( 6x + 1 - 6x + 1 )2
= 22 = 4
b ) x . ( 2x2 - 3 ) - x2 . ( 5x + 1 ) + x2
= 2x3 - 3x - 5x3 - x2 + x2
= ( 2x3 - 5x3 ) - 3x - ( x2 - x2 )
= - 3x3 - 3x
= - 3x . ( x2 + 1)
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
Bài 1:
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-2\left(36x^2-1\right)\)
\(=36x^2+72x+1+36x^2-72x+1-72x^2+2\)
\(=4\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
c) \(x\left(2x^3-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^4-3x-5x^3-x^2+x^2\)
\(=2x^4-5x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
a) Ta có : 3(x - 2)2 + (x - 1)3 = x3 + 29
=> 3(x2 - 4x + 4) + x3 - 3x2 + 3x - 1 = x3 + 29
=> 3x2 - 12x + 12 + x3 - 3x2 + 3x - 1 - x3 - 29 = 0
=> -9x - 18 = 0
=> -9x = 18
=> x = -2
Vậy x = -2