11:
a: \(\sqrt{x^2+4}=x-2\)(ĐKXĐ: \(x\in R\))
=>\(\left\{{}\begin{matrix}x>=2\\x^2-4x+4=x^2+4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=2\\-4x=0\end{matrix}\right.\Leftrightarrow x=0\left(loại\right)\)
b: \(\sqrt{x^2-10x+25}=3-19x\)(ĐKXĐ: \(x\in R\)
=>\(\left|x-5\right|=3-19x\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(x-5\right)^2=\left(-19x+3\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(19x-3-x+5\right)\left(19x-3+x-5\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(18x+2\right)\left(20x-8\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\left(nhận\right)\\x=\dfrac{2}{5}\left(loại\right)\end{matrix}\right.\)
9:
a: ĐKXĐ: \(\dfrac{-5}{-x-7}>=0\)
=>\(\dfrac{5}{x+7}>=0\)
=>x+7>0
=>x>-7
b: ĐKXĐ: \(x^2-3x+2>=0\)
=>(x-1)(x-2)>=0
TH1: \(\left\{{}\begin{matrix}x-1>=0\\x-2>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=1\\x>=2\end{matrix}\right.\)
=>x>=2
TH2: \(\left\{{}\begin{matrix}x-1< =0\\x-2< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =1\\x< =2\end{matrix}\right.\)
=>x<=1
c: ĐKXĐ: \(\dfrac{x+3}{5-x}>=0\)
=>\(\dfrac{x+3}{x-5}< =0\)
TH1: \(\left\{{}\begin{matrix}x+3>=0\\x-5< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-3\\x< 5\end{matrix}\right.\)
=>-3<=x<5
TH2: \(\left\{{}\begin{matrix}x+3< =0\\x-5>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>5\\x< =-3\end{matrix}\right.\)
=>Loại
d: ĐKXĐ: x^2-5x+6>0
=>(x-2)(x-3)>0
=>x>3 hoặc x<2
Bài 8
a) ĐKXĐ: -5x - 10 ≥ 0
⇔ -5x ≥ 10
⇔ x ≤ -2
b) ĐKXĐ: Với mọi x ∈ R vì:
x² - 2x + 1 = (x - 1)² ≥ 0 với mp8j x ∈ R
c) ĐKXĐ: Với mọi x ∈ R vì:
2x² + 4x + 5 = 2(x² + 2x + 1) + 3
= 2(x + 1)² + 3 > 0 với mọi x ∈ R
d) Ta có:
-x² + 4x - 4 = -(x² - 4x + 4)
= -(x - 2)² ≤ 0 với mọi x ∈ R
Vậy ĐKXD: x - 2 = 0 ⇔ x = 2
