Question

2. Escreva na forma ax? + bx +c=0 as
seguintes equações do 22 grau:

ALGUÉM ME AJUDA

EU DOU 15 PONTOS QUEM ME AJUDA POR FAVOR ​

Answer 1

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$\tt e)~\sf x^2-\dfrac{1}{3}=\dfrac{1}{6}x^2\\\sf 6\cdot x^2-\diagup\!\!\!6^2\cdot\dfrac{1}{\diagup\!\!\!3}=\diagup\!\!\!6\cdot\dfrac{1}{\diagup\!\!\!6}x^2\\\sf6x^2-2=x^2\\\sf 6x^2-x^2-2=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf 5x^2-2=0\checkmark}}}}$

$\tt f)~\sf\dfrac{x^2}{4}+\dfrac{1}{10}=\dfrac{x^2}{5}+\dfrac{x}{2}\cdot(20)\\\sf \diagup\!\!\!\!\!20^5\cdot\dfrac{x^2}{\diagup\!\!\!4}+\diagup\!\!\!\!\!20^2\cdot\dfrac{1}{\diagup\!\!\!\!\!10}=\diagup\!\!\!\!\!20^4\cdot\dfrac{x^2}{\diagup\!\!\!\!5}+\diagup\!\!\!\!\!20^{10}\cdot\dfrac{x}{\diagup\!\!\!2}\\\sf5x^2+2=4x^2+10x\\\sf5x^2-4x^2-10x+2=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf x^2-10x+2=0\checkmark}}}}$

$\tt g)~\sf x+6=\dfrac{4x}{x-2}\\\sf(x+6)\cdot(x-2)=4x\\\sf x^2-2x+6x-12-4x=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf x^2-12=0\checkmark}}}}$

$\tt h)~\sf\dfrac{2x}{x-3}=\dfrac{x+1}{x+3}\implies 2x\cdot(x+3)=(x-3)\cdot(x+1)\\\sf 2x^2+6x=x^2+x-3x-3\\\sf 2x^2-x^2+6x-x+3x-3=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf x^2+8x-3=0\checkmark}}}}$

$\tt i)~\sf\dfrac{x}{x-1}+\dfrac{1}{x+1}=\dfrac{x-3x^2}{x^2-1}\\\sf\dfrac{x}{x-1}+\dfrac{1}{x+1}=\dfrac{x-3x^2}{(x-1)\cdot(x+1)}$

$\sf\diagup\!\!\!\!(x\diagup\!\!\!\!-\diagup\!\!\!\!1)\cdot(x+1)\cdot\dfrac{x}{\diagup\!\!\!\!(x\diagup\!\!\!\!-\diagup\!\!\!\!1)}+(x-1)\cdot\diagup\!\!\!\!(x\diagup\!\!\!\!+\diagup\!\!\!\!1)\cdot\dfrac{1}{\diagup\!\!\!\!(x\diagup\!\!\!\!+\diagup\!\!\!\!1)}=\diagup\!\!\!\!(x\diagup\!\!\!\!-\diagup\!\!\!\!1)\cdot\diagup\!\!\!\!(x\diagup\!\!\!\!+1)\cdot\dfrac{x-3x^2}{\diagup\!\!\!\!(x\diagup\!\!\!\!-\diagup\!\!\!\!1)\cdot\diagup\!\!\!\!(x\diagup\!\!\!\!+\diagup\!\!\!\!1)}$

$\sf x\cdot(x+1)+x-1=x-3x^2\\\sf x^2+x+x-1=x-3x^2\\\sf x^2+3x^2+x+\diagup\!\!\!\!x-\diagup\!\!\!\!x-1=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf4x^2+x-1=0\checkmark}}}}$