ĐKXĐ: ...

\(y\left(y^2-5y+4\right)+y^2=\left(y^2-5y+4\right)\sqrt{x+1}+x+1\)

\(\Leftrightarrow\left(y^2-5y+4\right)\left(y-\sqrt{x+1}\right)+\left(y+\sqrt{x+1}\right)\left(y-\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\left(y-\sqrt{x+1}\right)\left[\left(y-2\right)^2+\sqrt{x+1}\right]=0\)

\(\Leftrightarrow y=\sqrt{x+1}\Rightarrow y^2=x+1\)

Thế xuống pt dưới:

\(2\sqrt{x^2-3x+3}+6x-7=\left(x+1\right)\left(x-1\right)^2+x\sqrt{3x-2}\)

\(\Leftrightarrow2\left(\sqrt{x^2-3x+3}-1\right)+x\left(x-\sqrt{3x-2}\right)=x^3-7x+6\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{\sqrt{x^2-3x+3}+1}+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=\left(x+3\right)\left(x^2-3x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\\\dfrac{2}{\sqrt{x^2-3x+3}+1}+\dfrac{x}{x+\sqrt{3x-2}}=x+3\left(1\right)\end{matrix}\right.\)

Xét (1) với \(x\ge\dfrac{3}{2}\):

\(\dfrac{2}{\sqrt{x^2-3x+3}+1}\le8-4\sqrt{3}< 1\)

\(\sqrt{3x-2}\ge0\Rightarrow\dfrac{x}{x+\sqrt{3x-2}}\le1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x^2-3x+3}+1}+\dfrac{x}{x+\sqrt{3x-2}}< 2\\x+3>2\end{matrix}\right.\) 

\(\Rightarrow\left(1\right)\) vô nghiệm

Bài 3: (d) cắt (d2) tại điểm C nằm trên trục tung

=>\(\begin{cases}a<>a^{\prime}\\ b=b^{\prime}\end{cases}\Rightarrow\begin{cases}m+3<>m-4\\ 2m-10=-2m-8\end{cases}\Rightarrow2m-10=-2m-8\)

=>2m+2m=-8+10

=>4m=2

=>m=0,5

Khi m=0,5 thì(d): y=(0,5+3)x+2*0,5-10=3,5x-9

Khi m=0,5 thì (d2): y=(0,5-4)x-2*0,5-8=-3,5x-9

Tọa độ A là:

\(\begin{cases}y=0\\ 3,5x-9=0\end{cases}\Rightarrow\begin{cases}y=0\\ 3,5x=9\end{cases}=>\begin{cases}y=0\\ x=\frac{9}{3,5}=\frac{18}{7}\end{cases}\)

Tọa độ B là:

\(\begin{cases}y=0\\ -3,5x-9=0\end{cases}\Rightarrow\begin{cases}y=0\\ -3,5x=9\end{cases}\Rightarrow\begin{cases}y=0\\ x=-\frac{18}{7}\end{cases}\)

B(-18/7;0); C(0;-9); A(18/7;0)

\(BC=\sqrt{\left(0+\frac{18}{7}\right)^2+\left(-9-0\right)^2}=\sqrt{\left(\frac{18}{7}\right)^2+9^2}=\sqrt{\frac{324}{49}+81}=\sqrt{\frac{4293}{49}}=\frac{\sqrt{4293}}{7}\)

\(AC=\sqrt{\left(\frac{18}{7}-0\right)^2+\left(0+9\right)^2}=\sqrt{\left(\frac{18}{7}\right)^2+9^2}=\frac{\sqrt{4293}}{7}\)

\(OA=\sqrt{\left(\frac{18}{7}-0\right)^2+\left(0-0\right)^2}=\frac{18}{7}\)

\(OB=\sqrt{\left(-\frac{18}{7}-0\right)^2+\left(0-0\right)^2}=\frac{18}{7}\)

Vì OA=OB và AC=BC

nên \(\frac{OA}{BC}=\frac{OB}{AC}\)

Câu 4:

Xét ΔOAI và ΔOBI có

OA=OB

AI=BI

OI chung

Do đó: ΔOAI=ΔOBI