\(\Rightarrow\left|a\right|\le1\),\(\left|b\right|\le1\),\(\left|c\right|\le1\)

\(\Rightarrow1-a\ge0\)tương tự 1-b,1-c............

\(\Rightarrow\left(1\right)\ge0\)

dấu = khi a=1b=0c=0 và hoán vị

Đang nổi cơn làm biếng mà nhìn thấy hệ còn buồn ngủ hơn:

a/ \(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-2x\right)\left(2x-y\right)=6\\x^2-2x-2\left(2x-y\right)=1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2-2x=a\\2x-y=b\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}ab=6\\a-2b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=6\\a=2b+1\end{matrix}\right.\)

\(\Rightarrow b\left(2b+1\right)=6\Leftrightarrow2b^2+b-6=0\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: ...

\(\Leftrightarrow\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}+\sqrt{2-\frac{1}{y}}-\sqrt{2-\frac{1}{x}}=0\)

\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\frac{1}{x}-\frac{1}{y}}{\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}}=0\)

\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{y-x}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)

\(\Leftrightarrow\frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}}+\frac{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}=0\)

\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\frac{1}{\sqrt{xy}}+\frac{\sqrt{y}+\sqrt{x}}{xy\left(\sqrt{2-\frac{1}{y}}+\sqrt{2-\frac{1}{x}}\right)}\right)=0\)

\(\Leftrightarrow\sqrt{y}=\sqrt{x}\Rightarrow x=y\)

Thay vào pt đầu:

\(\frac{1}{\sqrt{x}}+\sqrt{2-\frac{1}{x}}=2\)

\(\Leftrightarrow\frac{1}{x}+2-\frac{1}{x}+2\sqrt{\frac{2}{x}-\frac{1}{x^2}}=4\)

\(\Leftrightarrow\sqrt{\frac{2}{x}-\frac{1}{x^2}}=1\)

\(\Leftrightarrow\frac{2}{x}-\frac{1}{x^2}=1\)

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

c/ \(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2+x+y+xy=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\x^2+y^2=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=7\\\left(x+y\right)^2-2xy=10\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\) ta được:

\(\left\{{}\begin{matrix}a+b=7\\a^2-2b=10\end{matrix}\right.\) \(\Rightarrow a^2+2a-24=0\Rightarrow\left[{}\begin{matrix}a=4;b=3\\a=-6;b=13\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

a: Đặt |x-6|=a, |y+1|=b

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}2a+3b=5\\5a-4b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

=>|x-6|=1 và |y+1|=1

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

b: Đặt |x+y|=a, |x-y|=b

Theo đề, ta có: \(\left\{{}\begin{matrix}2a-b=19\\3a+2b=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{55}{7}\\b=-\dfrac{23}{7}\left(loại\right)\end{matrix}\right.\)

=>HPTVN

c: Đặt |x+y|=a, |x-y|=b

Theo đề ta có: \(\left\{{}\begin{matrix}4a+3b=8\\3a-5b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)

=>|x+y|=2 và x=y

=>|2x|=2 và x=y

=>x=y=1 hoặc x=y=-1

a.

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

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

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

\(\Leftrightarrow4x^3-3x+1=0\)

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

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

- Với \(x=-1\) thế vào pt đầu: \(1+y^2=\dfrac{1}{2}\Rightarrow y^2=-\dfrac{1}{2}\) (vô nghiệm)

- Với \(x=\dfrac{1}{2}\) thế vào pt đầu: \(\dfrac{1}{4}+y^2=\dfrac{1}{2}\Rightarrow y=\pm\dfrac{1}{2}\)

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

Dễ thấy x = 0 không phải nghiệm ta nhân tử mẫu phương trình đầu cho 3x thì được

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

Lấy (1) - (2) thì đơn giản rồi ha

b.

Trừ vế cho vế:

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

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

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

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

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

\(\Leftrightarrow x^2=y^2\) (ngoặc to hiển nhiên dương)

Thế vào pt đầu:

\(\left(1+x^2\right)^2\left(1+\dfrac{1}{x^4}\right)=8\)

Ta có: \(\left(1+x^2\right)^2\left(1+\dfrac{1}{x^4}\right)\ge4x^2.2\sqrt{1.\dfrac{1}{x^4}}=8\)

Đẳng thức xảy ra khi và chỉ khi \(x^2=1\)

Vậy nghiệm của hệ là \(x^2=y^2=1\Rightarrow x;y\)

Đặt S=x+y;P=xy giải ra :V

1,ĐK: \(x,y\ne-2\)

HPT<=> \(\left\{{}\begin{matrix}x\left(x+2\right)+y\left(y+2\right)=\left(x+2\right)\left(y+2\right)\left(1\right)\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x^2\left(x+2\right)^2+2xy\left(x+2\right)\left(y+2\right)+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)

=> \(2xy\left(x+2\right)\left(y+2\right)=0\)

<=>\(2xy=0\) (do x+2 và y+2 \(\ne0\))

<=> \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Tại x=0 thay vào (1) có: \(y\left(y+2\right)=2\left(y+2\right)\) <=> y= \(\pm2\) => y=2 (vì y khác -2)

Tại y=0 thay vào (1) có: \(x\left(x+2\right)=2\left(x+2\right)\) => x=2

Vậy HPT có 2 nghiệm duy nhất (2,0),(0,2)

2, ĐK: \(y\ne-1\)

HPT <=> \(\left\{{}\begin{matrix}x^2=2\left(x+3\right)\left(y+1\right)\left(1\right)\\\frac{3x^2}{y+1}=4-x\end{matrix}\right.\)

=> \(\frac{6\left(3+x\right)\left(y+1\right)}{y+1}=4-x\)

<=> 6(x+3)=4-x

<=> \(14=-7x\)

<=> \(x=-2\) thay vào (1) có \(4=2\left(y+1\right)\)

<=>y=1\(\)( tm)

Vậy hpt có một nghiệm duy nhất (-2,1)

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

PT (1) <=> \(\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)

<=> (x-y)(x+y+1)=0

<=>\(\left[{}\begin{matrix}x=y\\y=-x-1\end{matrix}\right.\)

Tại x=y thay vào (2) có \(y^2-y=y+3\) <=> \(y^2-2y-3=0\) <=> (y-3)(y+1)=0 <=> \(\left[{}\begin{matrix}y=3\\y=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Tại y=-1-x thay vào (2) có: \(x^2-x=-1-x+3\) <=> \(x^2=2\) <=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}y=-1-\sqrt{2}\\y=-1+\sqrt{2}\end{matrix}\right.\)

Vậy hpt có 4 nghiệm (3,3),(-1,-1), ( \(\sqrt{2},-1-\sqrt{2}\)),( \(-\sqrt{2},-1+\sqrt{2}\))

4,\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\left(2\right)\end{matrix}\right.\)(đk:\(x\ne0,y\ne0\))

<=> \(\left\{{}\begin{matrix}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=\frac{9}{2}\\\left(y+\frac{1}{y}\right)\left(x+\frac{1}{x}\right)=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=u\\y+\frac{1}{y}=v\end{matrix}\right.\)

Có \(\left\{{}\begin{matrix}u+v=\frac{9}{2}\\uv=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\v\left(\frac{9}{2}-v\right)=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left(v-\frac{5}{2}\right)\left(v-2\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left[{}\begin{matrix}v=\frac{5}{2}\\v=2\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\\\left[{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

Tại \(\left\{{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=\frac{5}{2}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)\left(y-\frac{1}{2}\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=2\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Tại \(\left\{{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\y+\frac{1}{y}=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x-2\right)\left(x-\frac{1}{2}\right)=0\\\left(y-1\right)^2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy hpt có 4 nghiệm (1,2),( \(1,\frac{1}{2}\)) ,( 2,1),(\(\frac{1}{2},1\)).

10.

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2xy-xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

@Nguyễn Việt Lâm @Lê Thị Thục Hiền @Akai Haruma @Trần Thanh Phương