\(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ =\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}\\ =\dfrac{x+2\sqrt{x}}{x-4}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Để \(A>1\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}-2}>1\Leftrightarrow\dfrac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}>0\Leftrightarrow2>0\left(LD\right)\)

\(\Leftrightarrow\sqrt{x}-2>0\Leftrightarrow x>4\left(tm\right)\)

Vậy \(x>4\) thì \(A>1\).

Bài 7: 

\(\widehat{C}=180^0-70^0-40^0=70^0=\widehat{B}\)

\(\left\{{}\begin{matrix}2y-3x=7\\y+4x=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8y-12x=28\\3y+12x=27\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}11y=55\\y+4x=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=5\\5+4x=9\end{matrix}\right.\)

                                                              \(\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=1\end{matrix}\right.\)

Vậy nghiệm hpt: \(\left(x;y\right)=\left(1;5\right)\)

15:

a: \(\text{Δ}=\left(m^2-m+2\right)^2-4m^2\)

=(m^2-m+2-2m)(m^2-m+2+2m)

=(m^2+m+2)(m^2-3m+2)

=(m-1)(m-2)(m^2+m+2)

Để phương trình co hai nghiệm phân biệt thì (m-1)(m-2)(m^2+m+2)>0

=>(m-1)(m-2)>0

=>m>2 hoặc m<1

b: x1+x2=m^2-m+2>0 với mọi m

x1*x2=m^2>0 vơi mọi m

=>Phương trình luôn có hai nghiệm dương phân biệt

a: Ta có: \(\left|x-0.6\right|< \dfrac{1}{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-0.6\ge0\\x-0.6< \dfrac{1}{3}\end{matrix}\right.\Leftrightarrow0.6\le x< \dfrac{14}{15}\)