\(=a^2xy+b^2xy-abx^2-aby^2\)

\(=ay\left(ax-by\right)-bx\left(ax-by\right)\)

\(=\left(ax-by\right)\left(ay-bx\right)\)

xy(a²+b²)-ab(x²+y²)

= xya²+xyb²-abx²-aby²

=(xya²-aby²)-(abx²-xyb²)

=ay(xa-by)-xb(xa-by)

=(xa-by)(ay-xb)

ĐKXĐ: \(xy\ne0;x\ne\pm y\)

\(\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{\dfrac{1}{y}+\dfrac{1}{x}}=\dfrac{5}{2}\\\dfrac{1}{y}-\dfrac{1}{x}+\dfrac{1}{\dfrac{1}{y}-\dfrac{1}{x}}=\dfrac{10}{3}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a+b+\dfrac{1}{a+b}=\dfrac{5}{2}\\b-a+\dfrac{1}{b-a}=\dfrac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-\dfrac{5}{2}\left(a+b\right)+1=0\\\left(b-a\right)^2-\dfrac{10}{3}\left(b-a\right)+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=2\\a+b=\dfrac{1}{2}\end{matrix}\right.\\\left[{}\begin{matrix}b-a=3\\b-a=\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a+b=2\\b-a=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{5}{2}\end{matrix}\right.\)

3 TH còn lại xét tương tự

Lấy pt trên trừ dưới ta được:

\(x^2+2y^2+3xy-x-3y-2=0\)

\(\Leftrightarrow x^2+\left(3y-1\right)x+2y^2-3y-2=0\)

Coi đây là pt bậc 2 ẩn x tham số y, ta có:

\(\Delta=\left(3y-1\right)^2-4\left(2y^2-3y-2\right)=\left(y+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3y+1+y+3}{2}=-y+2\\x=\dfrac{-3y+1-y-3}{2}=-2y-1\end{matrix}\right.\)

Thế vào pt đầu:

\(\Rightarrow\left[{}\begin{matrix}2\left(-y+2\right)^2+y^2+5y\left(-y+2\right)-y+2=0\\2\left(-2y-1\right)^2+y^2+5y\left(-2y-1\right)-y+2=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

Không tính ra được tiếp à thầy?

Từ pt đầu:

\(x^3-x^2y+2y^2-2xy=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}y=x\\y=\dfrac{x^2}{2}\end{matrix}\right.\)

- Với \(y=x\) thay vào pt dưới:

\(2\sqrt{x^2-2x+1}+\sqrt[3]{x^3-14}=x-2\)

\(\Leftrightarrow2\left|x-1\right|+\sqrt[3]{x^3-14}=x-2\)

TH1: \(x\ge1\)

\(\Rightarrow2x-2+\sqrt[3]{x^3-14}=x-2\)

\(\Leftrightarrow\sqrt[3]{x^3-14}=-x\)

\(\Leftrightarrow x^3-14=-x^3\)

\(\Leftrightarrow x^3=7\Rightarrow x=\sqrt[3]{7}\Rightarrow y=\sqrt[3]{7}\)

TH2: \(x< 1\)

\(\Rightarrow2-2x+\sqrt[3]{x^3-14}=x-2\)

\(\Leftrightarrow\sqrt[3]{x^3-14}=3x-4\)

\(\Leftrightarrow x^3-14=\left(3x-4\right)^3\)

\(\Leftrightarrow26x^3-108x^2+144x-50=0\)

Pt bậc 3 này nghiệm rất xấu (hay ko giải được theo chương trình phổ thông)

- Với \(y=\dfrac{x^2}{2}\), thay vào pt dưới:

\(2\sqrt[]{x^2-x^2+1}+\sqrt[3]{\left(\dfrac{x^2}{2}\right)^3-14}=x-2\)

\(\Leftrightarrow\sqrt[3]{\dfrac{x^6}{8}-14}=x-4\)

\(\Leftrightarrow\dfrac{x^6}{8}-14=\left(x-4\right)^3\)

Pt bậc 6 này thì càng ko giải được

\(\Rightarrow x^2+9x-1-2\sqrt{x^2+9x-1}-3=0\)

Đặt \(\sqrt{x^2+9x-1}=t\ge0\)

\(\Rightarrow t^2-2t-3=0\Rightarrow\left[{}\begin{matrix}t=-1< 0\left(loại\right)\\t=3\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+9x-1}=3\)

\(\Leftrightarrow x^2+9x-1=9\)

\(\Leftrightarrow x^2+9x-10=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\)

a) Để phương trình có nghiệm \(x_1,x_2\)

Thì \(\Delta'>0\)

\(\Leftrightarrow\left(m-2\right)^2-1.\left(2m-5\right)>0\)

\(\Leftrightarrow m^2-4m+4-2m+5>0\)

\(\Leftrightarrow m^2-6m+9>0\)

\(\Leftrightarrow\left(m-3\right)^2>0\)

\(\Leftrightarrow m\ne3\)

b)Với m khác 3. Theo hệ thức viet ta có

\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1.x_2=2m-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2m-4\left(1\right)\\x_1.x_2=2m-5\left(2\right)\end{matrix}\right.\)

Lấy (1) trừ (2) ta được

\(x_1+x_2-x_1.x_2=1\) không phụ thuộc vào m

ĐKXĐ: \(x\ne1;y\ne-1\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}-2x+2y=-4\\3x+2y=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{8}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+3y=4x-3\\14x+42y=15-9y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=-3\\14x+51y=15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-22\\y=\dfrac{19}{3}\end{matrix}\right.\)