\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)

\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

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

\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)

\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)

a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)

\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)

mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)

\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)

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

\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)

b) Tương tự câu a, ta có:

\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

c. Tương tự, ta có:

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

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

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

a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...

b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...

c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...

a: |2x+1|+|y-1|=4

mà 2x+1 lẻ

nên (|2x+1|;|y-1|)∈{(1;3);(3;1)}

=>(2x+1;y-1)∈{(1;3);(3;1);(-1;-3);(-3;-1);(1;-3);(-3;1);(-1;3);(3;-1)}

=>(2x;y)∈{(0;4);(2;2);(-2;-2);(-4;0);(0;-2);(-4;2);(-2;4);(2;0)}

=>(x;y)∈{(0;4);(1;2);(-1;-2);(-2;0);(0;-1);(-2;2);(-1;4);(1;0)}

b: \(y^2=3-\left|2x-3\right|\)

=>\(3-\left|2x-3\right|\ge0\)

=>|2x-3|<=3

mà 2x-3 lẻ

nên |2x-3|∈{1;3}

TH1: |2x-3|=1

=>\(y^2=3-1=2\)

mà y nguyên

nên y∈∅

TH2: |2x-3|=3

=>\(y^2=3-3=0\)

=>y=0(nhận)

|2x-3|=3

=>2x-3=3 hoặc 2x-3=-3

=>2x=6 hoặc 2x=0

=>x=3(nhận) hoặc x=0(nhận)

c: (x-3)(y-5)=-7

=>(x-3;y-5)∈{(1;-7);(-1;7);(7;-1);(-7;1)}

=>(x;y)∈{(4;-2);(2;12);(10;4);(-4;6)}

a: |2x+1|+|y-1|=4

mà 2x+1 lẻ

nên (|2x+1|;|y-1|)∈{(1;3);(3;1)}

=>(2x+1;y-1)∈{(1;3);(3;1);(-1;-3);(-3;-1);(1;-3);(-3;1);(-1;3);(3;-1)}

=>(2x;y)∈{(0;4);(2;2);(-2;-2);(-4;0);(0;-2);(-4;2);(-2;4);(2;0)}

=>(x;y)∈{(0;4);(1;2);(-1;-2);(-2;0);(0;-1);(-2;2);(-1;4);(1;0)}

b: \(y^2=3-\left|2x-3\right|\)

=>\(3-\left|2x-3\right|\ge0\)

=>|2x-3|<=3

mà 2x-3 lẻ

nên |2x-3|∈{1;3}

TH1: |2x-3|=1

=>\(y^2=3-1=2\)

mà y nguyên

nên y∈∅

TH2: |2x-3|=3

=>\(y^2=3-3=0\)

=>y=0(nhận)

|2x-3|=3

=>2x-3=3 hoặc 2x-3=-3

=>2x=6 hoặc 2x=0

=>x=3(nhận) hoặc x=0(nhận)

c: (x-3)(y-5)=-7

=>(x-3;y-5)∈{(1;-7);(-1;7);(7;-1);(-7;1)}

=>(x;y)∈{(4;-2);(2;12);(10;4);(-4;6)}

5x²+2y+y²-4x-40=0

△=(-4)²-4.5.(2y+y²-40)

△=16-40y-20y²+800

△=-(784+40y+20y²)

△=-(32y+8y+16y²+4y²+16+4+764)

△=-[(4y+4)²+(2y+2)²+764]<0

=>PHƯƠNG TRÌNH VÔ NGHIỆM.

3x + 9xy - 6y