Решите пожалуйста 7 и 8 задачт. Created by Lizochka5001. matematika-ru

Решите пожалуйста 7 и 8 задачт...

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$7)\; \; 2x^2-x-6=0\; ,\; D=1+48=49\; ,\; x_1=-\frac{3}{2}\; ,\; x_2=2\\\\2x^2-x-6=2(x+1,5)(x-2)=(2x+3)(x-2)\\\\\frac{7x}{2x^2-x-6}=\frac{7x}{(2x+3)(x-2)}=\frac{A}{2x+3}+\frac{B}{x-2}=\frac{A(x-2)+B(2x+3)}{(2x+3)(x-2)}\\\\7x=A(x-2)+B(2x+3)\\\\x=2\; \; \to \; \; 7\cdot 2=0+B(4+3)\; ,\; \; B=2\\\\x=-3/2\; \to \; \; 7\cdot (-\frac{3}{2})=A(-\frac{3}{2}-2)+0\; ,\; \; A=3\\\\\frac{7x}{2x^2-x-6}=\frac{3}{2x+3}+\frac{2}{x-2}$

$8)\; \; a=-3\\\\(\frac{a^3-64}{a^3-8a^2+16a}+\frac{4}{a-4})\cdot (a+4)^{-2}=\\\\=\Big (\frac{(a-4)(a^2+4a+16)}{a(a^2-8a+16)}+\frac{4}{a-4}\Big )\cdot \frac{1}{(a+4)^2}=\\\\=\Big (\frac{(a-4)(a^2+4a+16)}{a(a-4)^2}+\frac{4}{a-4}\Big )\cdot \frac{1}{(a+4)^2}=\\\\=\frac{a^2+4a+16 +4a}{a(a-4)}\cdot \frac{1}{(a-4)^2}=\frac{(a+4)^2}{a(a-4)}\cdot \frac{1}{(a+4)^2}=\frac{1}{a(a-4)}=\frac{1}{-3\cdot (-3-4)}=\frac{1}{21}$

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mishka19

$\frac{7x}{2x^2-x-6}=\frac{7x}{2x^2+3x-4x-6}=\frac{7x}{x(2x+3)-2(2x+3)}=\frac{7x}{(2x+3)(x-2)}\\ \\ \frac{7x}{(2x+3)(x-2)}=\frac{A}{2x+3}+\frac{B}{x-2} \\ \\ \frac{7x}{(2x+3)(x-2)}=\frac{A(x-2)+B(2x+3)}{(2x+3)(x-2)} \\ \\ 7x=A(x-2)+B(2x+3)\\ \\ 7x=Ax-2A+2Bx+3B\\ \\ 7x=(A+2B)x+(3B-2A)$

$\left \{ {{A+2B=7} \atop {3B-2A=0}} \right. \Leftrightarrow \left \{ {{2A+4B=14} \atop {3B-2A=0}} \right.\Leftrightarrow \left \{ {{A+2B=7} \atop {7B=14}} \right.\Leftrightarrow \left \{ {{A=7-2B} \atop {B=2}} \right.\Leftrightarrow\\ \\\Leftrightarrow\left \{ {{A=7-2\cdot2} \atop {B=2}} \right.\Leftrightarrow \left \{ {{A=3} \atop {B=2}} \right.$

$\frac{7x}{2x^2-x-6}=\frac{3}{2x+3}+\frac{2}{x-2}$

Ответ: $\frac{7x}{2x^2-x-6}=\frac{3}{2x+3}+\frac{2}{x-2}$

$(\frac{a^3-64}{a^3-8a^2+16a}+\frac{4}{a-4})\cdot(a+4)^{-2}=\frac{1}{a^2-4a}\\ \\ 1) \frac{a^3-64}{a^3-8a^2+16a}+\frac{4}{a-4}=\frac{a^3-4^3}{a(a^2-8a+16)}+\frac{4}{a-4}=\frac{(a-4)(a^2+a\cdot4+4^2)}{a(a^2-2\cdot a\cdot4+4^2)}+\frac{4}{a-4}=\\ \\ =\frac{(a-4)(a^2+4a+16)}{a(a-4)^2}+\frac{4}{a-4}=\frac{a^2+4a+16}{a(a-4)}+\frac{4}{a-4}^{(a}=\frac{a^2+4a+16+4a}{a(a-4)}=\\ \\ =\frac{a^2+8a+16}{a^2-4a}=\frac{a^2+2\cdot a\cdot4+4^2}{a^2-4a}=\frac{(a+4)^2}{a^2-4a}$

$2)\frac{(a+4)^2}{a^2-4a}\cdot(a+4)^{-2}=\frac{(a+4)^2}{a^2-4a}\cdot\frac{1}{(a+4)^{2}}= \frac{1}{a^2-4a}$

Ответ: $\frac{1}{a^2-4a}$

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