\(\Delta=\left(m+2\right)^2-4\cdot1\cdot2m\)

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

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

Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\)

=>\(\left(m-2\right)^2>0\)

=>\(m-2\ne0\)

=>\(m\ne2\)

Theo Vi-et, ta có:

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

\(x_1x_2+2\left(x_1+x_2\right)=2m+2\left(-m-2\right)=-4\)

=>Đây chính là hệ thức không phụ thuộc vào m

\(x^2-12x+4=0\)

Áp dụng định lý Vi-ét ta có :

\(\left\{{}\begin{matrix}x_1+x_2=12\\x_1.x_2=4\end{matrix}\right.\)

Theo đề bài ta có :

\(P=\dfrac{x_1^2+x_2^2}{\sqrt[]{x_1^2x_2}+\sqrt[]{x_1x_2^2}}\)

\(\Leftrightarrow P=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{\sqrt[]{x_1x_2}\left(x_1+x_2\right)}\)

\(\Leftrightarrow P=\dfrac{12^2-2.4}{\sqrt[]{4}.12}\)

\(\Leftrightarrow P=\dfrac{144-8}{2.12}\)

\(\Leftrightarrow P=\dfrac{136}{24}=\dfrac{17}{3}\)

\(A=\dfrac{x}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}-2}+\dfrac{2x-x\sqrt{x}-2}{x-3\sqrt{x}+2}\)

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

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

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

\(=\dfrac{2\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{2}{\sqrt{x}-1}\)

ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

\(A=\left(\dfrac{\sqrt{a}-2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}+2}{a-1}\right)\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)

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

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

\(=\dfrac{a-3\sqrt{a}+2-a-3\sqrt{a}-2}{\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{-6\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

\(=-\dfrac{6}{\sqrt{a}-1}\)

a) Thay m = 0 vào pt ta có:

\(x^2-2(0+1)x+0^2+0-1=0\)

\(\Leftrightarrow x^2-2x-1=0\)

\(\Leftrightarrow x^2-2x+1-2=0\)

\(\Leftrightarrow\left(x-1\right)^2-2=0\)

\(\Leftrightarrow\left(x-1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{matrix}\right.\)

Vậy ....

b) Để phương trình có 2 nghiệm phân biệt thì:

\(\Delta'>0\Leftrightarrow\left(m+1\right)^2-\left(m^2+m-1\right)>0\)

\(\Leftrightarrow m^2+2m+1-m^2-m+1>0\)

\(\Leftrightarrow m+2>0\Leftrightarrow m>-2\)

Theo viet:

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1.x_2=m^2+m-1\end{matrix}\right.\)

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=4\)

\(\Leftrightarrow\dfrac{x_2}{x_1x_2}+\dfrac{x_1}{x_1x_2}=4\)

\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=4\)

\(\Leftrightarrow\dfrac{2m+2}{m^2+m-1}=4\)(đk \(m^2+m-1\ne0\))

\(\Leftrightarrow\dfrac{2m+2}{m^2+m-1}-\dfrac{4m^2+4m-4}{m^2+m-1}=0\) (đk \(m^2+m-1\ne0\))

\(\Leftrightarrow\dfrac{-4m^2-2m+6}{m^2+m-1}=0\) (đk \(m^2+m-1\ne0\))

\(\Leftrightarrow-4m^2-2m+6=0\Leftrightarrow\left[{}\begin{matrix}m=1\left(tm\right)\\m=-\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)

Vậy 

a) m = 1

Phương trình tương đương:

x² - 2x = 0

⇔ x(x - 2) = 0

⇔ x = 0 hoặc x - 2 = 0

*) x - 2 = 0

⇔ x = 2

Vậy S = {0; 2}

b) Sửa đề: x₁(1 - x₂) + x₂(1 - x₁) = -2

∆ = [-(m + 1)]² - 4.(m - 1)

= m² + 2m + 1 - 4m + 4

= m² - 2m + 1 + 4

= (m - 1)² + 4 > 0 với mọi m ∈ R

Phương trình luôn có 2 nghiệm phân biệt với mọi m ∈ R

Theo Vi-ét, ta có:

x₁ + x₂ = m + 1

x₁x ₂= m - 1

x₁(1 - x₂) + x₂(1 - x₁) = -2

⇔ x₁ - x₁x₂ + x₂ - x₁x₂ = -2

⇔ x₁ + x₂ - 2x₁x₂ = -2

⇔ m + 1 - 2(m - 1) = -2

⇔ m + 1 - 2m + 2 = -2

⇔ -m + 3 = -2

⇔ -m = -2 - 3

⇔ -m = -5

⇔ m = 5

Vậy m = 5 thì phương trình có 2 nghiệm phân biệt thỏa mãn:

x₁(1 - x₂) + x₂(1 - x₁) = -2

m=2 mà

= 3 - √2 - 3√2 - 4(3 - √2)

= 3 - 4√2 - 12 + 4√2

= -9

1: Tọa độ A là:

\(\left\{{}\begin{matrix}2x+4=-x+1\\y=-x+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=-3\\y=-x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\left(-1\right)+1=2\end{matrix}\right.\)

vậy: A(-1;2)

2: O(0;0); A(-1;2); B(-1;-4)

\(OA=\sqrt{\left(-1-0\right)^2+\left(2-0\right)^2}=\sqrt{5}\)

\(OB=\sqrt{\left(-1-0\right)^2+\left(-4-0\right)^2}=\sqrt{17}\)

\(AB=\sqrt{\left(-1+1\right)^2+\left(-4-2\right)^2}=6\)

Xét ΔOAB có \(cosAOB=\dfrac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}=\dfrac{5+17-36}{2\cdot\sqrt{5}\cdot\sqrt{17}}=-\dfrac{7}{\sqrt{85}}\)

=>\(sinAOB=\sqrt{1-\left(-\dfrac{7}{\sqrt{85}}\right)^2}=\dfrac{6}{\sqrt{85}}\)

Diện tích tam giác AOB là:

\(S_{AOB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB\)

\(=\dfrac{1}{2}\cdot\dfrac{6}{\sqrt{85}}\cdot\sqrt{5}\cdot\sqrt{17}=3\)