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\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)
\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)
\(=-5x-27\)
\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y+z-x-y\right)^2\)
\(=z^2\)
a)
=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\)
=-5x-19
b)
=\(8x^3+y^3-8x^3+y^3\)
=\(2y^3\)
c)
=(x+y+z-x-y)\(^2\) +x+y
=\(z^2+x+y\)
hc tốt
\(A=\dfrac{2x}{x\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3}{x-y}\)
\(=\dfrac{2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{2x-2y+6x-3x-3y}{\left(x-y\right)\left(x+y\right)}=\dfrac{5\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{5}{x+y}\)
\(A=\left(x+y-1\right)3-\left(x-y+1\right)3+6x^2\left(y-1\right)\)
\(A=3\left(x+y-1-x+y-1\right)+6x^2\left(y-1\right)\)
\(A=3\left(2y-2\right)+6x^2\left(y-1\right)\)
\(A=3.2\left(y-1\right)+6x^2\left(y-1\right)\)
\(A=6\left(y-1\right)+6x^2\left(y-1\right)\)
\(A=6\left(y-1\right)\left(1+x^2\right)\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
\(x^2+6x-7\)
\(=x^2+x-7x-7\)
\(=x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left(x-7\right)\)
$x^4-12x^3+12x^2-12x+111$ tại $x=11$
$=11^4-12\cdot11^3+12\cdot11^2-12\cdot11+111$
$=14641-15972+1452-132+111$
$=100$
$(6x+1)^2+(6x-1)^2-2(1-6x)(6x-1)$
$=(6x+1)^2+(6x-1)^2+2(6x-1)^2$
$=(6x+1)^2+3(6x-1)^2$
$=36x^2+12x+1+3(36x^2-12x+1)$
$=36x^2+12x+1+108x^2-36x+3$
$=144x^2-24x+4$
$=4(36x^2-6x+1)$
b)$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$
$=3\cdot\dfrac{2^{32}-1}{2^2-1}$
$=3\cdot\dfrac{2^{32}-1}{3}$
$=2^{32}-1$