Phương trình có 2 nghiệm dương pb khi:

\(\left\{{}\begin{matrix}\Delta'=\left(m+1\right)^2-\left(2m-5\right)>0\\x_1+x_2=2\left(m+1\right)>0\\x_1x_2=2m-5>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m^2+6>0\\m>-1\\m>\dfrac{5}{2}\end{matrix}\right.\)

\(\Rightarrow m>\dfrac{5}{2}\)

Khi đó:

\(\left|\dfrac{1}{\sqrt{x_1}}-\dfrac{1}{\sqrt{x_2}}\right|=\sqrt{6}\Rightarrow\left|\dfrac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{x_1x_2}}\right|=\sqrt{6}\)

\(\Rightarrow\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)^2}{x_1x_2}=6\Rightarrow x_1+x_2-2\sqrt{x_1x_2}=6x_1x_2\)

\(\Rightarrow2\left(m+1\right)-2\sqrt{2m-5}=6\left(2m-5\right)\)

\(\Leftrightarrow5\left(2m-5\right)+2\sqrt{2m-5}-7=0\)

Đặt \(\sqrt{2m-5}=t>0\Rightarrow5t^2+2t-7=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{7}{5}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2m-5}=1\Rightarrow2m-5=1\)

\(\Rightarrow m=3\) (thỏa mãn)

BTVN Nâng cao thì tự làm hoặc là  ghi không biết vào đó

Chứ t chịu

Bài 4: 

a) Ta có: \(\dfrac{x}{13}=\dfrac{35}{91}\)

\(\Leftrightarrow\dfrac{x}{13}=\dfrac{5}{13}\)

hay x=5

Vậy: x=5

b) Ta có: \(\dfrac{9+x}{13-x}=\dfrac{5}{6}\)

\(\Leftrightarrow6\left(x+9\right)=5\left(13-x\right)\)

\(\Leftrightarrow6x+54=65-5x\)

\(\Leftrightarrow6x+5x=65-54\)

\(\Leftrightarrow11x=11\)

hay x=1

Vậy: S={1}

c) Ta có: \(x+x:5\cdot7.5+x:2\cdot9=315\)

\(\Leftrightarrow x+\dfrac{1}{5}x\cdot\dfrac{15}{2}+\dfrac{1}{2}x\cdot9=315\)

\(\Leftrightarrow7x=315\)

hay x=45

Vậy: x=45

câu 18

=6,46 m²

câu 19:

=150 l

Câu 1: A
Câu 2: A
Câu 3: B

5C

4B

1C

2C

3D