Question

Решить уравнения (желательно через ОДЗ), решение покороче
Алгебра 10 класс

Answer 1

Ответ:

1) $\emptyset$

2) $0$

Объяснение:

1)

ОДЗ

$2x+1\ge0\\\\2x\ge-1\ \ \ |:2\\\\x\ge-\frac{1}{2}\\\\\\4x+3\ge0\\\\4x\ge-3\ \ \ |:4\\\\x\ge-\frac{3}{4}\\\\x\in[-\frac{1}{2};+\infty)$

$\sqrt{2x+1}-\sqrt{4x+3}=1\ \ \ |()^2\\\\2x+1-2\sqrt{2x+1}\cdot \sqrt{4x+3}+4x+3=1\\\\6x+4-2\sqrt{(2x+1)(4x+3)}=1\\\\-2\sqrt{(2x+1)(4x+3)}=1-6x-4\\\\-2\sqrt{(2x+1)(4x+3)}=-6x-3\ \ \ |()^2\\\\4(2x+1)(4x+3)=36x^2+36x+9\\\\4(8x^2+6x+4x+3)=36x^2+36x+25\\\\32x^2+24x+16x+12=36x^2+36x+9\\\\32x^2+24x+16x+12-36x^2-36x-9=0\\\\-4x^2+4x+3=0$

$D=4^2-4\cdot(-4)\cdot3=16+48=64\\\\\sqrt{D}=\sqrt{64}=8\\\\x_1=\frac{-4-8}{2\cdot(-4)}=\frac{-12}{-8}=\frac{3}{2}=1,5\\\\x_2=\frac{-4+8}{2\cdot(-4)}=\frac{4}{-8}=-\frac{1}{2}=-0,5$

проверка

$\sqrt{2x+1}-\sqrt{4x+3}=\sqrt{2\cdot 1,5+1}-\sqrt{4\cdot 1,5+3}=\\\\\sqrt{3+1}-\sqrt{6+3}=\sqrt{4}-\sqrt{9}=2-3=-1 \neq 1$

$\sqrt{2x+1}-\sqrt{4x+3}=\sqrt{2\cdot (-0,5)+1}-\sqrt{4\cdot (-0,5)+3}=\\\\\sqrt{-1+1}-\sqrt{-2+3}=\sqrt{0}-\sqrt{1}=0-1=-1 \neq 1$

$x\in\emptyset$

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2)

ОДЗ

$1+x\ge0\\\\x\ge-1\\\\x\in[-1;+\infty)$

$\sqrt{1+x}=1-x\ \ \ |()^2\\\\1+x=1-2x+x^2\\\\1+x-1+2x-x^2=0\\\\-x^2+3x=0\\\\-x(x-3)=0\\\\x=0\ \ \ \ \ \ \ \ \ \ x-3=0\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=3$

проверка

$\sqrt{1+0}=\sqrt1=1\\\\1-x=1-0=1\\\\\\\sqrt{1+3}=\sqrt4=2\\\\1-x=1-3=-2 \neq 2$

$\underline {x=0}$

======================

1) 2-й способ

ОДЗ

$2x+1\ge0\\\\2x\ge-1\ \ \ |:2\\\\x\ge-\frac{1}{2}\\\\\\4x+3\ge0\\\\4x\ge-3\ \ \ |:4\\\\x\ge-\frac{3}{4}\\\\x\in[-\frac{1}{2};+\infty)$

$\sqrt{2x+1}-\sqrt{4x+3}=1\\\\\sqrt{2x+1}=1+\sqrt{4x+3}\ \ \ \ |()^2\\\\2x+1=1+2\sqrt{4x+3}+4x+3\\\\2x+1-1-4x-3=2\sqrt{4x+3}\\\\-2x-3=2\sqrt{4x+3}$

$x\in[-\frac{1}{2};+\infty)$

$-2x-3<0$

$2\sqrt{4x+3}>0$

$x\in \emptyset$