(-x^2*y)^3*1/2*x^2*y^3*(-2*x*y^2*z)^2
giup vs a
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\(\left(-x^3.z.y\right).\left(\dfrac{2}{3}.y.x^2\right)^2\)
\(=-x^3.z.y.\dfrac{4}{9}.y^2.x^4\)
\(=-\dfrac{4}{9}x^7.y^3.z\)
`(-x^3zy)(2/3yx^2)^2`
`=-4/9x^3zy.y^2x^4`
`=-4/9x^{3+4}.y^{1+2}z`
`=-4/9x^7y^3z`
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(T=\dfrac{2x^2-y^2}{2x^2+y^2}=\dfrac{2\left(2k\right)^2-\left(3k\right)^2}{2\left(2k\right)^2+\left(3k\right)^2}=\dfrac{8k^2-9k^2}{8k^2+9k^2}=\dfrac{-k^2}{17k^2}=\dfrac{-1}{17}\)
\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)
\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)
\(=-5x-27\)
\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y+z-x-y\right)^2\)
\(=z^2\)
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
\(\left(-x^2y\right)^3\cdot\dfrac{1}{2}\cdot x^2y^3\cdot\left(-2xy^2z\right)^2\\ =-x^6y^3\cdot\dfrac{1}{2}x^2y^3\cdot4x^2y^4z^2\\ =\left(-1\cdot\dfrac{1}{2}\cdot4\right)\cdot\left(x^6\cdot x^2\cdot x^2\right)\cdot\left(y^3\cdot y^3\cdot y^4\right)\cdot z^2\\ =-2x^{10}y^{10}z^2\)
bạn chụp ảnh sẽ dễ nhìn hơn :D