Question

(2х-3)(2х+3)(4х^2+9)=10х^2-82

Answer 1

Ответ:

Объяснение:

(4х^2-9)(4х^2+9)=10х^2-82

16х^4-81=10х^2-82

16х^4-10х^2=-1

2х^2(8х^2-5)=-1

8х^2-5=-1

8х^2=4

х^2=4/8

х^2=1/2

х= √0,5. х=-√0,5

Answer 2

Ответ:

$\pm \frac{\sqrt{2}}{2}, \quad \pm \frac{\sqrt{2}}{4} \quad ;$

Объяснение:

$(2x-3)(2x+3)(4x^{2}+9)=10x^{2}-82;$

$((2x)^{2}-3^{2})(4x^{2}+9)=10x^{2}-82;$

$(4x^{2}-9)(4x^{2}+9)=10x^{2}-82;$

$(4x^{2})^{2}-9^{2}-10x^{2}+82=0;$

$16x^{4}-81-10x^{2}+82=0;$

$16x^{4}-10x^{2}+1=0;$

Введём замену:

$t=x^{2};$

Перепишем уравнение с учётом замены:

$16t^{2}-10t+1=0;$

$D=b^{2}-4ac;$

$D=(-10)^{2}-4 \cdot 16\cdot 1=100-64=36;$

$t_{1}=\frac{-b+\sqrt{D}}{2a};$

$t_{1}=\frac{-(-10)+\sqrt{36}}{2 \cdot 16}=\frac{10+6}{32}=\frac{16}{32}=\frac{1}{2};$

$t_{2}=\frac{-b-\sqrt{D}}{2a};$

$t_{2}=\frac{-(-10)-\sqrt{36}}{2 \cdot 16}=\frac{10-6}{32}=\frac{4}{32}=\frac{1}{8};$

Вернёмся к замене:

$x^{2}=\frac{1}{2} \quad \vee \quad x^{2}=\frac{1}{8};$

$x= \pm \sqrt{\frac{1}{2}} \quad \vee \quad x= \pm \sqrt{\frac{1}{8}} \quad ;$

$x= \pm \frac{\sqrt{1}}{\sqrt{2}} \quad \vee \quad x= \pm \frac{\sqrt{1}}{\sqrt{8}} \quad ;$

$x= \pm \frac{1}{\sqrt{2}} \quad \vee \quad x= \pm \frac{1}{2\sqrt{2}} \quad ;$

$x= \pm \frac{\sqrt{2}}{2} \quad \vee \quad x= \pm \frac{\sqrt{2}}{4} \quad ;$

Answer 3

Let's simplify the expression step by step:

(2x - 3)(2x + 3)(4x^2 + 9)

First, notice that the first two terms form a difference of squares, which can be simplified as follows:

(2x - 3)(2x + 3) = (2x)^2 - (3)^2 = 4x^2 - 9

Now, we have:

(4x^2 - 9)(4x^2 + 9)

This is a difference of squares as well:

(4x^2)^2 - (9)^2 = 16x^4 - 81

So, the simplified expression is:

16x^4 - 81

Now, if you want to find the difference between this expression and 10x^2 - 82:

16x^4 - 81 - (10x^2 - 82)

Now, distribute the negative sign to both terms in the second expression:

16x^4 - 81 - 10x^2 + 82

Combine like terms:

16x^4 - 10x^2 + 1

So, the difference between 16x^4 - 81 and 10x^2 - 82 is:

16x^4 - 10x^2 + 1