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a) \(6xy+4x-9y-7=0\)
\(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)
\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)
\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)
Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)
Tự làm típ
\(A=x^3+y^3+xy\)
\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)
\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))
\(A=x^2+y^2\)
Áp dụng bất đẳng thức Bunhiakovxky ta có :
\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)
\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)
Hay \(x^3+y^3+xy\ge\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
CM : với a,b > 0 thì \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b};\frac{\left(a+b\right)^2}{4}\ge ab\)
Dấu " = " xảy ra \(\Leftrightarrow\)a = b
Ta có : P = \(\frac{5}{x^2+y^2}+\frac{3}{xy}=\left(\frac{5}{x^2+y^2}+\frac{5}{2xy}\right)+\frac{1}{2xy}=5.\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}\)
\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}=\frac{4}{9}\)
\(xy\le\frac{\left(x+y\right)^2}{4}\Rightarrow\frac{1}{2xy}\ge\frac{2}{\left(x+y\right)^2}=\frac{2}{9}\)
\(\Rightarrow P\ge5.\frac{4}{9}+\frac{2}{9}=\frac{22}{9}\)
Dấu " = "xảy ra \(\Leftrightarrow\)x = y = 1,5
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Áp dụng bất đẳng thức Svacxo và bất đẳng thức \(\frac{1}{4ab}\ge\frac{1}{\left(a+b\right)^2}\)ta có :
\(Q=\frac{2}{x^2+y^2}+\frac{2}{2xy}+\frac{4}{2xy}=2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{8}{4xy}\)
\(\ge2\frac{\left(1+1\right)^2}{\left(x+y\right)^2}+\frac{8}{\left(x+y\right)^2}=\frac{2.4}{2^2}+\frac{8}{2^2}=\frac{16}{4}=4\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=1\)
Vậy min Q = 4 khi x = y = 1
\(B=\frac{x^3}{y+1}+\frac{y^3}{1+x}=\frac{\left(x^4+y^4\right)+\left(x^3+y^3\right)}{xy+x+y+1}\)
\(=\frac{\left(x^4+y^4\right)+\left(x+y\right)\left(x^2+y^2-xy\right)}{x+y+2}=\frac{\left(x^4+y^4\right)+\left(x+y\right)\left(x^2+y^2-1\right)}{x+y+2}\)
Áp dụng BĐT cô si với các số dương x2 ; y2 ; x4 ; y4 ta được :
\(B\ge\frac{2x^2y^2+\left(x+y\right)\left(2xy-1\right)}{x+y+2}=\frac{2+\left(x+y\right)}{x+y+2}=1\)
Dấu ''='' xảy ra khi \(\Leftrightarrow x=y=1\)
Mình tự làm tận 1h nên hơi dài 1 tí nhưng chắc chắn đúng đó :))
Ta có: x2 + y2 + xy .- 3x - 3y + 3 = 0
=>( x2 - 2x + 1) - x + ( y2 - 2y + 1) - y + xy + 1 = 0
=> (x-1)2 + (y-1)2 + ( -x + -y + xy +1) = 0
=> (x-1)2 + (y-1)2 + [(-x+ xy) + (-y+1)] = 0
=> (x-1)2 + (y-1)2 + [ x(y-1) - (y-1)] = 0
=> (x-1)2 + (y-1)2 + (x-1)(y-1) = 0
=> (x-1)2 + 2.1/2.(x-1)(y-1) + (1/2)2.(y-1)2 + 3/4.(y-1)2 = 0
=> [x-1+1/2(y-1) ]2 + 3/4.(y-1)2 = 0
Vì: [x-1+1/2(y-1) ]2 >= 0 với mọi x;y thuộc R
3/4.(y-1)2 >= 0 với mọi y thuộc R
=> (x-1+1/2y -1/2 = 0) và ( y-1 = 0)
=> (x = 1/2 -1/2y+1) và (y=1)
=> x = y =1
Chỗ này thay giá trị vào biểu thức rồi chứng minh = cách chỉ ra các cơ số của từng lũy thừa là số nguyên là xong.
