nguyên tử oxi

nguyên tố O

phân tử O₂

srr bạn vì lúc nãy mik đọc lộn đề    

\(a,A=\left(x+y\right)^2-9x^2=\left(x+y\right)^2-\left(3x\right)^2=\left(x+y-3x\right).\left(x+y+3x\right)=\left(y-2x\right).\left(4x+y\right)\) \(b,A=9-x^2-2xy-y^2=3^2-\left(x^2+2xy+y^2\right)=3^2-\left(x+y\right)^2=\left(3-x-y\right).\left(3+x+y\right)\)

Khối lượng của 0,25 mol phân tử Fe₃O₄ là:

m_Fe3O4= n.M = 0,25. (56.3 + 16.4)=58 (g)

\(a,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right).\left(x-y+z\right)\)

\(b,x^3+y^3+2x^2-2xy+2y^2=\left(x^3+y^3\right)+2\left(x^2-xy+y^2\right)=\left(x+y\right).\left(x^2-2xy+y^2\right)+2.\left(x^2-xy+y^2\right)=\left(x^2-xy+y^2\right).\left(x+y+2\right)\)

\(a,\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)-\left(4x\right)^2=\left(x^2+4-4x\right).\left(x^2+4+4x\right)=\left(x-2\right)^2.\left(x+2\right)^2\)

\(b,\left(x+3\right)^3-8x^3=\left(x+3\right)^3-\left(2x\right)^3=\left(x+3-2x\right).\left[x^2+\left(x+3\right).2x+\left(2x\right)^2\right]=\left(3-x\right).\left(x^2+2x^2+6x+4x^2\right)\)

\(c,\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=\left(4x^2-3x-18-4x^2-3x\right).\left(4x^2-3x-18+4x^2+3x\right)=\left(-6x-18\right).\left(8x^2-18\right)\)

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a, Ta có: a//b ,  a//c => b//c

b, Do a//b => \(\widehat{A}+\widehat{AOb}=180^o\left(2gócTCP\right)\)

                 =>\(120^o+\widehat{AOb}=180^o\)

                 => \(\widehat{AOb}=60^o\left(1\right)\)

Do b//c => \(\widehat{bOB}+\widehat{OBc}=180^o\left(2gócTCP\right)\)

             => \(\widehat{bOB}+120^o=180^o\)

             => \(\widehat{bOB}=60^o\left(2\right)\)

Từ (1) và (2)=> \(\widehat{AOb}=\widehat{bOB}\)hay Ob là tia pg cuả \(\widehat{AOB}\left(dpcm\right)\)

đáp án: a,b  nhé 

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