Tham khảo: 

\(x=\dfrac{1}{a}.\sqrt{\dfrac{2a}{b}-1}\Rightarrow ax=\sqrt{\dfrac{2a}{b}-1}\)

\(\Rightarrow\left\{{}\begin{matrix}1+ax=\dfrac{\sqrt{2a-b}+\sqrt{b}}{\sqrt{b}}\\1-ax=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1-ax}{1+ax}=\dfrac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}+\sqrt{2a-b}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2\left(b-a\right)}\)

Lại có:

\(\dfrac{1+bx}{1-bx}=\dfrac{a+\sqrt{2ab-b^2}}{a-\sqrt{2ab-b^2}}=\dfrac{a^2-\left(2ab-b^2\right)}{\left(a-\sqrt{2ab-b^2}\right)^2}=\dfrac{\left(a-b\right)^2}{\left(a-\sqrt{2ab-b^2}\right)^2}\)

\(\Rightarrow\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{b-a}{a-\sqrt{2ab-b^2}}\)

\(\Rightarrow A=\dfrac{1-ax}{1+ax}.\sqrt{\dfrac{1+bx}{1-bx}}=\dfrac{\left(\sqrt{b}-\sqrt{2a-b}\right)^2}{2a-2\sqrt{2ab-b^2}}=\dfrac{2a-2\sqrt{2ab-b^2}}{2a-2\sqrt{2ab-b^2}}=1\)

\(Q=\frac{1+\text{ax}}{1-\text{ax}}\sqrt{\frac{1-bx}{1+bx}}\)

Ta có: \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\Rightarrow\text{ax}=\sqrt{\frac{2a-b}{b}}\Rightarrow1+\text{ax}=1+\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}}\)

\(1-\text{ax}=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\)

\(\Rightarrow\frac{1+\text{ax}}{1-\text{ax}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}-\sqrt{2a-b}}=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2b-2a}\left(1\right)\)

 \(bx=\frac{b}{a}\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}\left(2a-b\right)}{a}\Rightarrow\hept{\begin{cases}1-bx=\frac{a-\sqrt{b\left(2a-b\right)}}{a}\\1+bx=\frac{a+\sqrt{b\left(2a-b\right)}}{a}\end{cases}}\)

\(\Rightarrow\frac{1-bx}{1+bx}=\frac{a-\sqrt{b\left(2a-b\right)}}{a+\sqrt{b\left(2a-b\right)}}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{a^2-2ab+b^2}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{\left(a-b\right)^2}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow Q=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2\left(b-a\right)}.\frac{a-\sqrt{b\left(2a-b\right)}}{a-b}=\frac{\text{[}2a+2\sqrt{b\left(2a-b\right)}\text{]}\left(a-b\sqrt{2a-b}\right)}{2\left(a-b\right)^2}\)

\(\Rightarrow\frac{2\left[a^2-b\left(2a-b\right)\right]}{2\left(a-b\right)^2}=\frac{2\left(a^2-2ab+b^2\right)}{a\left(a-b\right)^2}=1\)

`a) 7x^2 - 2x + 3 = 0`

`(a = 7; b = -2; c = 3)`

`Δ = b^2 - 4ac = (-2)^2 - 4.7.3 = -80 < 0`

`=>` phương trình vô nghiệm

`b) 6x^2 + x + 5 = 0`

`(a = 6;b = 1;c = 5)`

`Δ = b^2 - 4ac = 1^2 - 4.6.5 = -119 < 0`

`=>` phương trình vô nghiệm

`c) 6x^2 + x - 5 = 0`

`(a = 6;b=1;c=-5)`

`Δ = b^2 - 4ac = 1^2 - 4.6.(-5) = 121 > 0`

`=>` phương trình có 2 nghiệm phân biệt

`x_1 = (-b + sqrt{Δ})/(2a) = (-1+ sqrt{121})/(2.6) = (-1+11)/12 = 10/12 = 5/6`

`x_2 = (-b - sqrt{Δ})/(2a) = (-1- sqrt{121})/(2.6) = (-1-11)/12 = -12/12 = -1`

Vậy phương trình có 1 nghiệm `x_1 = 5/6; x_2 = -1`

ủa, mấy bài đó tương tự như ct mà:

\(7x^2-2x+3=0\) \(\left\{{}\begin{matrix}a=7\\b=-2\\c=3\end{matrix}\right.\)

\(\Delta=b^2-4ac=\left(-2\right)^2-4.7.3=-80\)

Vì \(\Delta< 0\) \(\Rightarrow\) pt vô nghiệm

a)

`7x^2 -2x+3=0`

có \(\Delta=b^2-4ac=\left(-2\right)^2-4\cdot7\cdot3=-80< 0\)

=> phương trình vô nghiệm

b)

`6x^2 +x+5=0`

có \(\Delta=b^2-4ac=1^2-4\cdot6\cdot5=-119< 0\)

=> phương trình vô nghiệm

c)

`6x^2 +x-5=0`

có \(\Delta=b^2-4ac=1^2-4\cdot6\cdot\left(-5\right)=121>0\)

\(=>x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{121}}{2\cdot6}=\dfrac{5}{6}\)

\(=>x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{121}}{2\cdot6}=-1\)