a)
[tex]\dfrac{(x-1)(x+2)}{x^2+4x-5}=0\\\\x^2+4x-5 \neq 0 \\\\\Delta=4^2-4\cdot 1\cdot (-5) = 16+20=36\\\sqrt{\Delta}=6\\\\x_1=\dfrac{-4-6}{2}=\dfrac{-10}2=-5\\\\x_2=\dfrac{-4+6}2=\dfrac22=1\\\\D: x\in \mathbb{R}\setminus\{-5; 1\}[/tex]
[tex](x-1)(x+2)=0\\\\x-1=0 \wedge x+2=0\\\\x=1 \notin D\:\: \boxed{x=-2 \in D}[/tex]
d)
[tex]|2x-3|=1\\\\2x-3=1 \wedge -2x+3=1\\2x=4 \wedge -2x=-2\\\boxed{x=2 \wedge x=1}[/tex]
g)
[tex]|2x+3|\leq 5\\\\2x+3\leq 5 \wedge -2x-3\leq 5\\2x\leq 2 \wedge -2x \leq 8\\x\leq 1 \wedge x\geq -4\\\\\boxed{x\in\langle-4;1\rangle}[/tex]
h)
[tex]\left|\dfrac13x+2\right|\geq 1\\\\\dfrac13x+2\geq 1 \wedge -\dfrac13x-2\geq 1\\\\\dfrac13x\geq -1 \wedge -\dfrac13x \geq 3\\\\x\geq -3 \wedge x\leq -9\\\\\boxed{x\in(-\infty; -9\rangle\cup\langle-3;\infty)}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
a)
[tex]\dfrac{(x-1)(x+2)}{x^2+4x-5}=0\\\\x^2+4x-5 \neq 0 \\\\\Delta=4^2-4\cdot 1\cdot (-5) = 16+20=36\\\sqrt{\Delta}=6\\\\x_1=\dfrac{-4-6}{2}=\dfrac{-10}2=-5\\\\x_2=\dfrac{-4+6}2=\dfrac22=1\\\\D: x\in \mathbb{R}\setminus\{-5; 1\}[/tex]
[tex](x-1)(x+2)=0\\\\x-1=0 \wedge x+2=0\\\\x=1 \notin D\:\: \boxed{x=-2 \in D}[/tex]
d)
[tex]|2x-3|=1\\\\2x-3=1 \wedge -2x+3=1\\2x=4 \wedge -2x=-2\\\boxed{x=2 \wedge x=1}[/tex]
g)
[tex]|2x+3|\leq 5\\\\2x+3\leq 5 \wedge -2x-3\leq 5\\2x\leq 2 \wedge -2x \leq 8\\x\leq 1 \wedge x\geq -4\\\\\boxed{x\in\langle-4;1\rangle}[/tex]
h)
[tex]\left|\dfrac13x+2\right|\geq 1\\\\\dfrac13x+2\geq 1 \wedge -\dfrac13x-2\geq 1\\\\\dfrac13x\geq -1 \wedge -\dfrac13x \geq 3\\\\x\geq -3 \wedge x\leq -9\\\\\boxed{x\in(-\infty; -9\rangle\cup\langle-3;\infty)}[/tex]