Math, asked by Anonymous, 3 months ago

Don't Spam........Answer with Explanations......​

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Answered by khashrul
7

Answer:

Option (D) None of the these

Step-by-step explanation:

[3x^2y^3z^4 . \frac{4}{9} (x^2 + y) z^3][\frac{-3}{2} x^{2} y^{3} z^{4} (y + z + 3x)]

= -2x^4y^6z^{11} (x^2 + y)(y + z + 3x)

= -2x^4y^6z^{11} (x^2y + zx^2 + 3x^3 + y^2 + yz + 3xy)

= -2x^4y^6z^{11} (3x^3 + x^2y + zx^2 + 3xy + y^2 + yz)

= -6x^7y^6z^{11} -2x^6y^7z^{11} -2x^6y^6z^{12} -6x^5y^7z^{11} -2x^4y^8z^{11} -2x^4y^7z^{12}

Option (D) None of the these

Answered by Anonymous
133

♣ Qᴜᴇꜱᴛɪᴏɴ : Find the product of \sf{\left[3 x^{2} y^{3} z^{4} \cdot \dfrac{4}{9}\left(x^{2}+y\right) z^{3}\right]} and \sf{\left[\dfrac{-3}{2}\:x^2\:y^3\:z^4\left(y+z+3\:x\right)\right]}

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♣ ᴀɴꜱᴡᴇʀ :

\sf{-6x^7y^6z^{11}-2x^6y^7z^{11}-2x^6y^6z^{12}-6x^5y^7z^{11}-2x^4y^8z^{11}-2x^4y^7z^{12}}

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

\sf{\displaystyle\left[3\:x^2\:y^3\:z^4\:\cdot \frac{4}{9}\left(x^2+y\right)\:z^3\right]\left[\frac{-3}{2}\:x^2\:y^3\:z^4\left(y+z+3\:x\right)\right]}

\sf{\displaystyle=3y^3z^4\frac{4}{9}\left(x^2+y\right)z^3\frac{-3}{2}x^{2+2}y^3z^4\left(y+z+3x\right)}

\sf{\displaystyle=3y^3z^4\frac{4}{9}\left(x^2+y\right)z^3\frac{-3}{2}x^4y^3z^4\left(y+z+3x\right)}

\sf{\displaystyle=3z^4\frac{4}{9}\left(x^2+y\right)z^3\frac{-3}{2}x^4y^{3+3}z^4\left(y+z+3x\right)}

\sf{\displaystyle=3z^4\frac{4}{9}\left(x^2+y\right)z^3\frac{-3}{2}x^4y^6z^4\left(y+z+3x\right)}

\sf{\displaystyle=3\cdot \frac{4}{9}\left(x^2+y\right)\frac{-3}{2}x^4y^6z^{4+3+4}\left(y+z+3x\right)}

\sf{\displaystyle=3\cdot \frac{4}{9}\left(x^2+y\right)\frac{-3}{2}x^4y^6z^{11}\left(y+z+3x\right)}

\sf{\displaystyle=\frac{4\left(-3\right)\cdot \:3\left(x^2+y\right)x^4y^6z^{11}\left(y+z+3x\right)}{9\cdot \:2}}

\sf{=\dfrac{-36x^4y^6z^{11}\left(x^2+y\right)\left(3x+y+z\right)}{18}}

\sf{\displaystyle=-\frac{36\left(x^2+y\right)x^4y^6z^{11}\left(y+z+3x\right)}{18}}

\sf{=-2x^4y^6z^{11}\left(x^2+y\right)\left(3x+y+z\right)}

\sf{=-2x^4y^6z^{11}(x^2\cdot \:3x+x^2y+x^2z+y\cdot \:3x+yy+yz)}

\sf{=-2x^4y^6z^{11}(3x^2x+x^2y+x^2z+3xy+yy+yz)}

\sf{=-2x^4y^6z^{11}((3x^3+x^2y+x^2z+3xy+y^2+yz)}

\sf{=\left(-2x^4y^6z^{11}\right)\cdot \:3x^3+\left(-2x^4y^6z^{11}\right)x^2y+\left(-2x^4y^6z^{11}\right)x^2z+\left(-2x^4y^6z^{11}\right)\cdot \:3xy}

\sf{+\left(-2x^4y^6z^{11}\right)y^2+\left(-2x^4y^6z^{11}\right)yz}

\sf{=-2\cdot \:3x^4x^3y^6z^{11}-2x^4x^2y^6yz^{11}-2x^4x^2y^6z^{11}z-2\cdot \:3x^4xy^6yz^{11}-2x^4y^6y^2z^{11}-2x^4y^6yz^{11}z}

\sf{=-6x^7y^6z^{11}-2x^6y^7z^{11}-2x^6y^6z^{12}-6x^5y^7z^{11}-2x^4y^8z^{11}-2x^4y^7z^{12}}

Hence you go with Option (D) : "None of these"

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