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\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)
\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)
\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)
\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)
Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)
\(\Leftrightarrow A\ne0\forall x;y\)
2x2 + 2y2 + 3xy - x + y + 1 = 0
2x2 + 2y2 + 4xy - xy - x + y + 1 = 0
(2x2 + 2y2 + 4xy) + (-xy - x) + (y + 1) = 0
2(x + y)2 - x(y + 1) + (y + 1) = 0
2(x + y)2 + (y + 1)(1 - x) = 0
Do (x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 + (y + 1)(1 - x) = 0 \(\Leftrightarrow\) (y + 1)(1 - x) = 0
\(\Rightarrow y+1=0;1-x=0\)
*) y + 1 = 0
y = -1
*) 1 - x = 0
x = 1
Với x = 1; y = -1, ta có:
B = [1 + (-1)]2018 + (1 - 2)2018 + (-1 - 1)2018
= 1 + 22018
\(M=4\left(x-1\right)\left(x+1\right)-5x\left(x-2\right)+x^2\)
\(=4x^2-4-5x^2+10x+x^2\)
\(=10x-4\)
\(M=\left(y^2+2\right)\left(y-4\right)-\left(2y^2+1\right)\left(\dfrac{1}{2}y-2\right)\)
\(=\left(y^2+2\right)\left(y-4\right)-\dfrac{1}{2}\left(2y^2+1\right)\left(y-4\right)\)
\(=\left(y-4\right)\left(y^2+2-y^2-\dfrac{1}{2}\right)\)
\(=\dfrac{3}{2}y-6\)
c)
\(C=\left(3-2x\right)\left(x-2\right)-4\left(x-1\right)\left(x-3\right)-\left(x-2\right)\left(x+2\right)\)
= 3x - 6 - 2x2 + 4x - 4x2 + 12x + 4x - 12 - x2 + 4
= - 7x2 + 23x - 14
a) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)
\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)
\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)
\(Q=\left(x-y-2x-4y\right)^2\)
\(Q=\left(-x-5y\right)^2\)
b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)
\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)
\(A=\left[\left(xy+2\right)-2\right]^3\)
\(A=\left(xy+2-2\right)^3\)
\(A=\left(xy\right)^3\)
\(A=x^3y^3\)
c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)
\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)
\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)
\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)
\(=0\)
a: =(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2
=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2
b: =(xy+2-2)^3=(xy)^3=x^3y^3
c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)
=24x+2x^3-2x^3-24x
=0
a)
(x-2y)2 >= 0 V x,y
(y-2018)>=0 V y
=> P=(ghi lại đề) >= 0
vậy GTNN của p bằng 0
dấu "=" xảy ra (=) \(\hept{\begin{cases}x-2y=0\\y-2018=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\y=2018\end{cases}}\left(=\right)\hept{\begin{cases}y=2018\\x=4036\end{cases}}\)
b) (x+y-3)4 >= 0 V x,y
(x-2y)2 >= V x,y
=> Q=(ghi lại đề) >= 2018
vậy GTNN của Q bằng 2018
dấu "=" xảy ra (=) \(\hept{\begin{cases}x+y-3=0\\x-2y=0\end{cases}}\left(=\right)\hept{\begin{cases}x=2y\\3y=3\end{cases}}\left(=\right)\hept{\begin{cases}y=1\\x=2\end{cases}}\)
c)
(2x + 1/6)4>= 0 V x
=> N=(ghi lại đề) >= -2
vậy GTNN của N bằng -2
dấu "=" xảy ra (=) 2x+1/6=0
(=) 2x=-16
(=) x=-1/12
#Học-tốt
Menhera-Kun chữ V rồi gạch đó là j