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
 

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

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

1. Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) với \(a=x^3+3xy^2,b=y^3+3x^2y\) (a;b > 0)

(Bất đẳng thức này a;b > 0 mới dùng được)

\(A\ge\frac{4}{x^3+3xy^2+y^3+3x^2y}=\frac{4}{\left(x+y\right)^3}\ge\frac{4}{1^3}=4\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x^3+3xy^2=y^3+3x^2y\\x+y=1\end{cases}\Leftrightarrow\hept{\begin{cases}x^3-3x^2y+3xy^2-y^3=0\\x+y=1\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)^3=0\\x+y=1\end{cases}}\Leftrightarrow x=y=\frac{1}{2}\)