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Rút gọn biểu thức
a).x(2x2-3)-x2(5x+1)+x2
b).3x(x-2)-5x(1-x)-8(x2-3)
c).1/2x2(6x-3)-x(x2+1/2)+1/2(x+4)
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=4\)
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=\left(2x^3-5x^3\right)+\left(x^2-x^2\right)-3x\)
\(=-3x^3-3x\)
\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)
\(=-11x+24\)
$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2$
$=2^2$
$=4$
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)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
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) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
a) \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
b) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
$(6x+1)^2-(6x-1)-2(1+6x)(6x-1)$
$=36x^2+12x+1-6x+1-2(6x+1)(6x-1)$
$=36x^2+6x+2-2(36x^2-1)$
$=36x^2+6x+2-72x^2+2$
$=-36x^2+6x+4$
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)$3x(x-2)-5x(1-x)-8(x^2-3)$
$=3x^2-6x-5x+5x^2-8x^2+24$
$=3x^2+5x^2-8x^2-11x+24$
$=-11x+24$
$=24-11x$
\(a,x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
\(b,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
=.= hok tốt!!
a)
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2=x\left(2x^2-3-5x^2-x+x\right)\)
=\(x\left(-3x^2-3\right)=-3x\left(x^2+1\right)\)
b)
Nhân ra:
\(3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
Chúc bạn học tốt!