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$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
b)$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=2x^3-5x^3-3x$
$=-3x^3-3x$
$=-3x(x^2+1)$
c)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
b)$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=2x^3-5x^3-3x$
$=-3x^3-3x$
$=-3x(x^2+1)$
c)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
(6x+1)2+(6x-1)2-2(1+6x)(6x-1)=(6x+1+1-6x)2=4
x(2x2-3)-x2(5x+1)+x2=2x3-3x-5x3-x2= -3x3-x2-3x
3x(x-2)-5x(1-x)-8(x2-3)=3x2-6x-5x+5x2-8x2+24= -11x+24
chưa chắc là đúng đâu nhé
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+12x+1+36x^2-12x+1-72x^2+2\)
\(=4\)
c) \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
$(6x+1)^2-(6x-1)-2(1+6x)(6x-1)$
$=36x^2+12x+1-6x+1-2(6x+1)(6x-1)$
$=36x^2+6x+2-2(36x^2-1)$
$=36x^2+6x+2-72x^2+2$
$=-36x^2+6x+4$
b)$x(2x^2-3)-x^2(5x+1)+x^2$
$=2x^3-3x-5x^3-x^2+x^2$
$=2x^3-5x^3-3x$
$=-3x^3-3x$
$=-3x(x^2+1)$
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$
d)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
a) (6x+1)2 + (6x-1)2 - 2(1+6x)(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
= -3x3 - 3x = -3x(x2 -1) = -3x(x-1)(x+1)
c) 3x(x-2) - 5x(1-x) - 8(x2 -3)
= 3x2 - 6x - 5x + 5x2 -8x2 + 24
= -11x +24
a) (6x+1)2 + (6x-1)2 - 2(1+6x)(6x-1)
= (6x+1+6x-1)2
=144x2
b) x(2x2 -3) - x2(5x+1) +x2
=2x3 - 3x - 5x3 -x2+x2
=-3x3-3x
=-3x(x2+1)
c) 3(22+1)(24+1)(28+1)(216+1)
= (22-1)(22+1)(24+1)(28+1)(216+1)
= (24-1)(24+1)(28+1)(216+1)
= (28-1)(28+1)(216+1)
= (216-1)(216+1)
= 232 -1
d) 3x(x-2) - 5x(1-x) - 8(x2 -3)
= 3x2-6x - 5x + 5x2 - 8x2 +24
= -11x +24
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)$3(2^2-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3(2^2-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3(2^2-1)\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3(2^2-1)\cdot\dfrac{2^{32}-1}{2^4-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2+1}$
$=\dfrac{3(2^{32}-1)}{5}$
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=4\)
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=\left(2x^3-5x^3\right)+\left(x^2-x^2\right)-3x\)
\(=-3x^3-3x\)
\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)
\(=-11x+24\)