как решить вот это ?. Created by tupoyfg. algebra-ru

как решить вот это ?...

mukus13

Verified answer

$\sqrt{ \frac{(2x-3)^3}{2x^2-9x+7} } - \sqrt{ \frac{(x-1)^3}{4x^2-20x+21} } =0$

$\left \{ {{\frac{(x-1)^3}{4x^2-20x+21} \geq 0} \atop {(\sqrt{ \frac{(2x-3)^3}{2x^2-9x+7} })^2 = (\sqrt{ \frac{(x-1)^3}{4x^2-20x+21} })^2 } \right.$

${\frac{(x-1)^3}{4(x-3.5)((x-1.5)} \geq 0}$

- + - +
---------[1]------------(1.5)-----------(3.5)---------------
//////////////// /////////////////////

$x$ ∈ $[1;1.5)$ ∪ $(3.5;+$ ∞ $)$

$\sqrt{ \frac{(2x-3)^3}{2x^2-9x+7} } = \sqrt{ \frac{(x-1)^3}{4x^2-20x+21} }$

$( \sqrt{ \frac{(2x-3)^3}{2x^2-9x+7} })^2= (\sqrt{ \frac{(x-1)^3}{4x^2-20x+21} } )^2$

$\frac{(2x-3)^3}{2x^2-9x+7} ={ \frac{(x-1)^3}{4x^2-20x+21} }$

${2x^2-9x+7} =0$
$D=(-9)^2-4*2*7=25$
$x_1= \frac{9+5}{4}=3.5$
$x_2= \frac{9-5}{4}=1$

$4x^2-20x+21=0$
$D_1= (\frac{b}{2})^2-ac=(-10)^2-4*21=16$
$x_1= \frac{- \frac{b}{2}+ \sqrt{D_1} }{a} = \frac{10+4}{4} =3.5$
$x_2= \frac{- \frac{b}{2}- \sqrt{D_1} }{a} = \frac{10-4}{4} =1.5$

$\frac{(2x-3)^3}{2(x-1)(x-3.5)} ={ \frac{(x-1)^3}{4(x-1.5)(x-3.5)} }$

ОДЗ:
$x \neq 1$
$x \neq 3.5$
$x \neq 1.5$

$(2x-3)^3*4(x-1.5)(x-3.5)=(x-1)^3*2(x-1)(x-3.5)$

$(2x-3)^3*4(x-1.5)(x-3.5)-(x-1)^3*2(x-1)(x-3.5)=0$

$2(x-3.5)((2x-3)^3*2(x-1.5)-(x-1)^3(x-1))=0$

$2(x-3.5)((2x-3)^4-(x-1)^4)=0$

$(x-3.5)((2x-3)^2-(x-1)^2)((2x-3)^2+(x-1)^2))=0$

$(x-3.5)(2x-3-x+1)(2x-3+x-1)(4x^2+9-12x+x^2-2x+1)=0$

$(x-3.5)(x-2)(3x-4)(5x^2-14x+10)=0$

$x-3.5=0$ или $x-2=0$ или $3x-4=0$ или $5 x^{2} -14x+10=0$

$x=3.5$ или $x=2$ или $x=1 \frac{1}{3}$ или $D=(-14)^2-4*5*10\ \textless \ 0$

∅ ∅ ∅

Ответ: $1 \frac{1}{3}$

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