Question

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Answer 1

Объяснение:

1) ax=8

x=$\frac{8}{a}$

2)ax+2=a

$ax=a-2\\x=\frac{a-2}{a} \\x = 1 - \frac{2}{a}$

3)

$(b^{2} -2b)x=b^{2} -4\\x=\frac{b^{2} -4}{(b^{2} -2b)} \\x=\frac{(b -2)(b+2)}{b(b -2)} x=\frac{b+2}{b} x=1+ \frac{2}{b} \\$

4)

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

5)

$2(a-2x)=ax+3\\2a - 4x = ax + 3\\2a - 3 = ax + 4x\\2a - 3 = x(a + 4)\\x = \frac{2a - 3}{a + 4}$

6) $(a+1)x^{2} -2x + 1 -a = 0\\$

Дискриминант

$x_{1,2} = \frac{-b (+-) \sqrt{b^{2}-4ac } }{2a} \\x_{1} = \frac{2 + \sqrt{4-4(1-a)(a+1) } }{2(a+1)} x_{1} = \frac{2 + \sqrt{4-4(1-a^{2} ) } }{2(a+1)} x_{1} = \frac{2 + \sqrt{4a^{2} } }{2(a+1)} x_{1} = \frac{2 +2a }{2(a+1)} x_{1} = \frac{2(1 +a) }{2(a+1)} x_{1} = 1\\x_{2} = \frac{2 - \sqrt{4-4(1-a)(a+1) } }{2(a+1)} x_{2} = \frac{2 - \sqrt{4a^{2} } }{2(a+1)} x_{2} = \frac{2(1 -a) }{2(a+1)} x_{2} = \frac{1 -a }{a+1}$

7)

$x^{2} -(a-1)x^{2} +a -2a^{2} =0\\x^{2} (1-(a-1)) + a -2a^{2} =0\\x^{2} (-a) + a -2a^{2} =0x^{2} (-a) = 2a^{2} -a\\x^{2} = -\frac{2a^{2} -a}{a} \\x^{2} = -(2a -1) \\x^{2} = -2a + 1\\x = \sqrt{1- 2a}$

8)

$(a^{2} +7a - 8)x = a^{2} +16a+64\\x = \frac{a^{2} +16a+64}{a^{2} +7a - 8} \\x = \frac{(a +8)^{2}}{a^{2} +7a - 8} \\$