Chương 3: PHƯƠNG TRÌNH, HỆ PHƯƠNG TRÌNH

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

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

Xét \(pt\left(1\right)\Leftrightarrow2x^2+y^2-3xy-4x+3y+2=0\)

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

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

*)\(y=x-1\) thay vao \(pt(2)\) :

\(pt\Leftrightarrow\sqrt{x^2-x+4}=2\Leftrightarrow x^2-x=0\)

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

*)\(y=2x-2\) thay vao \(pt(2)\):

\(pt\Leftrightarrow\sqrt{x^2-2x+5}+\sqrt{x-1}=2\)

\(\Leftrightarrow\dfrac{x^2-2x+1}{\sqrt{x^2-2x+5}+2}+\sqrt{x-1}=0\)

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

\(\Leftrightarrow x=1\)\(\Leftrightarrow y=0\)

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

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

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

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

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

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

2) \(\left\{{}\begin{matrix}x^2+y^2=5\\x^4-x^2y^2+y^4=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\\left(x^2+y^2\right)^2-3x^2y^2=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\\left(5\right)^2-3x^2y^2=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\-3x^2y^2=-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\x^2y^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\\left(5-y^2\right)y^2=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\-y^4+5y^2-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\-y^4+5y^2-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=5-y^2\\\left[{}\begin{matrix}y^2=1\\y^2=4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=5-y^2\\y^2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=5-y^2\\y^2=4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2=4\\y^2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x^2=1\\y^2=4\end{matrix}\right.\end{matrix}\right.\)

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

vậy \(S=\left\{\left(2;1\right),\left(2;-1\right),\left(-2;1\right),\left(-2;-1\right),\left(1;2\right),\left(1;-2\right),\left(-1;2\right),\left(-1;-2\right)\right\}\)

4) \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^5-1=-y^5\\x^9-x^4+y^9-y^4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x^5-1=-y^5\\y^5-1=-x^5\end{matrix}\right.\\x^4\left(x^5-1\right)+y^4\left(y^5-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\x^4\left(-y^5\right)+y^4\left(-x^5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\-x^4y^4\left(x+y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^5+y^5=1\\\left[{}\begin{matrix}x=0\\y=0\\x+y=0\end{matrix}\right.\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-y\\x^5-x^5=1\end{matrix}\right.\end{matrix}\right.\)

vậy \(S=\left\{\left(1;0\right),\left(0;1\right)\right\}\)

từ pt (2) -> xy=2-x(x+1)-y(y+1)=2-4=-2.

từ pt (1) ->x²+y²-4+x+y=0. Thay xy=-2 vào ta được:

x²+y²+2xy+x+y=0<=>x+y=0 hoặc x+y=-1

Nếu x+y=0->x=-y ->y²=2 ->y=\(\pm\sqrt{2}\)->x=\(\mp\sqrt{2}\)

Nếu x+y=-1->\(\left[{}\begin{matrix}x=1;y=-2\\x=-2;y=1\end{matrix}\right.\)

Vậy (x;y)\(\in\left\{\left(\sqrt{2};-\sqrt{2}\right);\left(-\sqrt{2};\sqrt{2}\right);\left(1;-2\right);\left(2;-1\right)\right\}\)

\(\Leftrightarrow\left(x^2+12+8x\right)\left(x^2+12+7x\right)=420x^2\)

\(\Leftrightarrow\left(x^2+12\right)^2+15x\left(x^2+12\right)+56x^2-420x^2=0\)

\(\Leftrightarrow\left(x^2+12+28x\right)\left(x^2+12-13x\right)=0\)

=>(x-12)(x-1)(x²+28x+196-184)=0

hay \(x\in\left\{12;1;-14+2\sqrt{46};-14-2\sqrt{46}\right\}\)

\(\dfrac{1}{x-3}=\dfrac{x^2-3x+5}{x^2-x-6}\)

Suy ra: \(x^2-3x+5=x+2\)

=>x²-4x+3=0

=>(x-3)*(x-1)=0

=>x=1(nhận) hoặc x=3(loại)

\(\dfrac{1}{x-3}\)=\(\dfrac{x^2-3x+5}{x^2-x-6}\)

suy ra \(x\)²-3\(x\)+5=\(x\)=2

Lời giải: https://hoc24.vn/hoi-dap/question/905017.html