Задания 6,7,8 Заранее говорю спасибо )

Question 1

Задания 6,7,8
Заранее говорю спасибо )

Question 2

ШЕСТОЙ НОМЕРОК

$\frac{6}{x^4-6x^2+9}+1=\frac{7}{x^2-3}\\\frac{7}{x^2-3}-\frac{6}{x^4-6x^2+9}=1\\\frac{7}{x^2-3}*\frac{x^4-6x^2+9}{1}-\frac{6}{x^4-6x^2+9}*\frac{x^4-6x^2+9}{1}=1*\frac{x^4-6x^2+9}{1}\\7(x^2-3)-6=x^4-6x^2+9\\7x^2-21=x^4-6x^2+15\\x^4-13x^2+36=0\\zamena:x^2=a,a\ \textgreater \ 0\\a^2-13a+36=0\\\left[\begin{array}{ccc}a_1+a_2=13\\a_1*a_2=36\end{array}\right\to\left[\begin{array}{ccc}a_1=9\\a_2=4\end{array}\right\to\left[\begin{array}{ccc}x^2=9\\x^2=4\end{array}\right\to\left[\begin{array}{ccc}x=б3\\x=б2\end{array}\right$

$1-\frac{15}{x^4-8x^3+16x^2}=\frac{2}{x^2-4x}\\\frac{15}{x^4-8x^3+16x^2}+\frac{2}{x^2-4x}=1\\\frac{15}{x^4-8x^3+16x^2}*\frac{x^4-8x^3+16x^2}{1}+\frac{2}{x^2-4x}*\frac{x^4-8x^3+16x^2}{1}=1*\frac{x^4-8x^3+16x^2}{1}\\15+2(x^2-4x)=x^4-8x^3+16x^2\\x^4-8x^3+14x^2+8x-15=0\\x^4-x^2-8x^3+8x+15x^2-15=0\\x^2(x^2-1)-8x(x^2-1)+15(x^2-1)=0\\(x^2-1)(x^2-8x+15)=0\\x^2-1=0/x^2-8x+15=0\\Otvet:\left[\begin{array}{ccc}x_{1,2}=б1\\\left[\begin{array}{ccc}x_3=3\\x_4=5\end{array}\right\end{array}\right$

СЕДЬМОЙ НОМЕРОК

$\frac{8}{x^3+5}-\frac{1}{x^3+2}=1\\zamena:x^3+2=a,a\neq0\to\left[\begin{array}{ccc}x^3\neq-2\\x^3\neq-5\end{array}\right\to\left[\begin{array}{ccc}x\neq\sqrt[3]{-2}\\x\neq\sqrt[3]{-5}\end{array}\right\\\frac{8}{a+3}-\frac{1}{a}=1\\\frac{8a}{a+3}*\frac{a+3}{1}=(a+1)*\frac{a+3}{1}\\8a=(a+1)(a+3)=a^2+4a+3\\a^2-4a+3=0\\\left[\begin{array}{ccc}a_1=1\\a_2=3\end{array}\right\to\left[\begin{array}{ccc}x^3+2=1\\x^3+2=3\end{array}\right\to\left[\begin{array}{ccc}x_1=\sqrt[3]{-1}\\x_2=1\end{array}\right$

$\frac{12}{x^3+5}-\frac{4}{x^3+3}=1\\zamena:x^3+3=a,a\neq0\to\left[\begin{array}{ccc}x^3\neq-3\\x^3\neq-5\end{array}\right\to\left[\begin{array}{ccc}x\neq\sqrt[3]{-3}\\x\neq\sqrt[3]{-5}\end{array}\right\\\frac{12}{a+2}-\frac{4}{a}=1\\\frac{12a}{a+2}*\frac{a+2}{1}=(a+4)*\frac{a+2}{1}\\12a=(a+4)(a+2)=a^2+6a+8\\a^2-6a+8=0\\\left[\begin{array}{ccc}a_1=2\\a_2=4\end{array}\right\to\left[\begin{array}{ccc}x^3+3=2\\x^3+3=4\end{array}\right\to\left[\begin{array}{ccc}x_1=\sqrt[3]{-1}\\x_2=1\end{array}\right$

ВОСЬМОЙ НОМЕРОК

$\frac{x-3}{x^2+4x+9}+\frac{x^2+4x+9}{x-3}=-2\\zamena:\left[\begin{array}{ccc}x-3=a\\x^2+4x+9=b\end{array}\right\\-2=\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\to a^2+b^2=-2ab\to a^2+2ab+b^2=0\to \\(a+b)^2=0\to a+b=0\to(x-3)+(x^2+4x+9)=0\\x^2+5x+6=0\\\left[\begin{array}{ccc}x_1=-2\\x_2=-3\end{array}\right$

$\frac{x-5}{x^2+7x+3}+\frac{x^2+7x+3}{x-5}=2\\zamena:\left[\begin{array}{ccc}x-5=a\\x^2+7x+3=b\end{array}\right\\2=\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\to a^2+b^2=2ab\to a^2-2ab+b^2=0\to\\(a-b)^2=0\to a-b=0\to(x-5)-(x^2+7x+3)=0\\x-5-x^2-7x-3=0\\x^2+6x+8=0\\\left[\begin{array}{ccc}x_1=-4\\x_2=-2\end{array}\right$

Question 3

Спасибо, действительно помогли мне))))