Question

[Nguyễn Phúc Trường An]

 
                                       mọi người giải giúp em với ạ

Answer 1

[YangSu]

\(2/\)

\(a,\) Rút gọn

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

\(=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(b,\) Để \(A>1\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>1\)

 \(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}-1>0\) 

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

\(\Leftrightarrow\sqrt{x}-\sqrt{x}+2>0\)

\(\Leftrightarrow2>0\left(LD\right)\)

Vậy với mọi giá trị x thì \(A>1\)

\(3/\)

\(x^2-2\left(m+1\right)x+m^2+3m-1=0\)

\(\Delta=b^2-4ac=\left[-2\left(m+1\right)\right]^2-4\left(m^2+3m-1\right)\)

                      \(=4\left(m^2+2m+1\right)-4\left(m^2+3m-1\right)\)

                       \(=4m^2+8m+4-4m^2-12m+4\)

                       \(=-4m+8\)

Để pt có 2 nghiệm \(x_1,x_2\) thì \(\Delta>0\Leftrightarrow-4m+8>0\Leftrightarrow-4m>-8\Leftrightarrow m< 2\)

Theo Vi-ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2m+2\\x_1x_2=\dfrac{c}{a}=m^2+3m-1\end{matrix}\right.\)

Ta có : \(x_1^2+x_2^2=10\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-10=0\)

\(\Leftrightarrow\left(2m+2\right)^2-2\left(m^2+3m-1\right)-10=0\)

\(\Leftrightarrow4m^2+8m+4-2m^2-6m+2-10=0\)

\(\Leftrightarrow2m^2+2m-4=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=1\left(tmdk\right)\\m=-2\left(tmdk\right)\end{matrix}\right.\)

Vậy \(m=1,m=-2\) thì thỏa mãn đề bài.

Answer 2

[YangSu]

3b bạn tự vẽ hình nha.