A = \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

A = \(\left(\dfrac{x}{\left(x+7\right)\left(x-7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x\left(x+7\right)}+\dfrac{x}{7-x}\)

A = \(\left(\dfrac{x^2-\left(x-7\right)^2}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{2x-7}{x\left(x+7\right)}-\dfrac{x}{x-7}\)

A = \(\left(\dfrac{x^2-\left(x^2-14x+49\right)}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{\left(2x-7\right)\left(x-7\right)-\left(x^3+7x^2\right)}{\left(x+7\right)\left(x-7\right)x}\)

A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}:\dfrac{-x^3-5x^2-21x+49}{\left(x+7\right)\left(x-7\right)x}\)

A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}.\dfrac{\left(x+7\right)\left(x-7\right)x}{-x^3-5x^2-21x+49}\)

A = \(\dfrac{14x-49}{-x^3-5x^2-21x+49}\)

\(\Delta\) = 5² - 4(m - 2) = 25 - 4m + 8 = 33 - 4m

phương trình có 2 nghiệm phân biệt

\(\Leftrightarrow\) \(\Delta\) > 0 \(\Leftrightarrow\) 33 - 4m > 0 \(\Leftrightarrow\) - 4m > - 33 \(\Leftrightarrow\) m < \(\dfrac{33}{4}\)

phương trình có 2 nghiệm dương \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x_1+x_2>0\\x_1.x_2>0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}5>0\\m-2>0\end{matrix}\right.\) \(\Leftrightarrow\) m > 2

ta có : \(2\left(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}\right)\) = 3 \(\Leftrightarrow\) \(2\left(\dfrac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{x_1.x_2}}\right)\) = 3

\(\Leftrightarrow\) \(\dfrac{2\left(\sqrt{x_1}+\sqrt{x_2}\right)}{\sqrt{x_1.x_2}}\) = 3 \(\Leftrightarrow\) \(2\left(\sqrt{x_1}+\sqrt{x_2}\right)\) = \(3\sqrt{x_1.x_2}\)

\(\Leftrightarrow\) \(2\sqrt{x_1}\) + \(2\sqrt{x_2}\) = \(3\sqrt{x_1.x_2}\) \(\Leftrightarrow\) \(\left(2\sqrt{x_1}+2\sqrt{x_2}\right)^2\) = \(\left(3\sqrt{x_1.x_2}\right)^2\)

\(\Leftrightarrow\) 4x₁ + 8\(\sqrt{x_1.x_2}\) + 4x₂ = 9x₁.x₂ \(\Leftrightarrow\) 4(x₁ + x₂) + 8\(\sqrt{x_1.x_2}\) = 9x₁.x₂

áp dụng hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1.x_2=m-2\end{matrix}\right.\)

thay vào ta có : 20 + 8\(\sqrt{m-2}\) = 9(m-2)

\(\Leftrightarrow\) 20 + 8\(\sqrt{m-2}\) = 9m - 18 \(\Leftrightarrow\) 9m - 38 = 8\(\sqrt{m-2}\)

\(\Leftrightarrow\) (9m - 38)² = 64 (m - 2) (vì m - 2 > 0)

\(\Leftrightarrow\) 81m² - 684m + 1444 = 64m - 128

\(\Leftrightarrow\) 81m² - 748m + 1572 = 0

giải phương trình ta được m = 6 ; m = \(\dfrac{262}{81}\) (đều thỏa mảng điều kiện)

vậy m = 6 ; m = \(\dfrac{262}{81}\) là thỏa mãng điều kiện bài toán

a) \(\Delta\) = (2m + 1)² - 4 (m² + m - 6)

= 4m² + 4m +1 - 4m² - 4m + 24 = 25 > 0

\(\Rightarrow\) phương trình có 2 nghiệm phân biệt \(\forall\)m

phương trình (1) có 2 nghiệm âm \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x_1+x_2< 0\\x_1.x_2>0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2m+1< 0\\m^2+m-6>0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}m< \dfrac{-1}{2}\\m< -3\end{matrix}\right.\) vậy m < - 3

hoặc \(\left\{{}\begin{matrix}m< \dfrac{-1}{2}\\m>2\end{matrix}\right.\) m \(\in\) \(\varnothing\)

vậy m < - 3 thì phương trình (1) có 2 nghiệm âm