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a) 6x^2y^5 (-2) x^3y^2z b) (1/5 x^3y^4) (10/9xyz) bài 2 A= -x^2-5yz+z^2 B= 7yz-z^2+5x^2 a) tính A+B b) tính A-B
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*Lời giản
Bài 1
`a) 6x^2y^5 (-2) x^3y^2z`
`= (-2 . 6) (x^2 . x^3) (y^5 . y^2) z`
`= -12 x^5 y^7 z`
`b) (1/5 x^3y^4) (10/9 xyz)`
`= (1/5 . 10/9) (x^3 . x) (y^4 . y) z`
`= 2/9 x^4 y^5 z`
Bài 2
`a) A + B = -x^2 – 5yz + z^2 + 7yz – z^2 + 5x^2`
`-> A + B = (-x^2 + 5x^2) + (-5yz + 7yz) + (z^2 – z^2)`
`-> A + B = 4x^2 + 2yz`
`b) A – B = -x^2 – 5yz + z^2 – 7yz + z^2 – 5x^2`
`-> A – B = (-x^2 – 5x^2) + (-5yz – 7yz) + (z^2 + z^2)`
`-> A – B = -6x^2 – 12yz + 2z^2`
Giải thích các bước giải:
Bài 1:
a.Ta có:
$6x^2y^5(-2)x^3y^2z$
$=-6x^2y^5\cdot \:2x^3y^2z$
$=-6\cdot -2\cdot x^2\cdot x^3\cdot y^5\cdot y^2\cdot z$
$=-12x^5y^7z$
b.Ta có:
$(\dfrac15x^3y^4)(\dfrac{10}{9}xyz)$
$=\dfrac15x^3y^4\cdot \dfrac{10}{9}xyz$
$=\dfrac15\cdot \dfrac{10}9\cdot x^3\cdot x\cdot y^4\cdot y\cdot z$
$=\dfrac29x^4y^5z$
Bài 2:
a.Ta có:
$A+B=(-x^2-5yz+z^2)+(7yz-z^2+5x^2)$
$\to A+B=-x^2-5yz+z^2+7yz-z^2+5x^2$
$\to A+B=z^2-z^2-5yz+7yz+4x^2$
$\to A+B=2yz+4x^2$
Ta có:
$A-B=(-x^2-5yz+z^2)-(7yz-z^2+5x^2)$
$\to A-B=-x^2-5yz+z^2-7yz+z^2-5x^2$
$\to A-B=z^2+z^2-5yz-7yz-x^2-5x^2$
$\to A-B=2z^2-12yz-6x^2$