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a: \(VT=x^2+2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1\)
\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1>0\forall x,y\)
c: \(VT=x^2-6xy+9y^2+4x^2-4x+1+y^2-2y+1+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1>0\forall x,y\)
a)
\(x^2+xy+y^2+1=\left(x^2+2x\times\frac{y}{2}+\left(\frac{y}{2}\right)^2\right)+\frac{3y^2}{4}+1\)
\(=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\ge0+0+1=1\)
mà\(1>0\Rightarrow x^2+xy+y^2+1>0\)với mọi \(x\)và\(y\)
b)
\(x^2+5y^2+2x-4xy-10y+14\)
\(=\left[x^2+2x\left(1-2y\right)+\left(1-2y\right)^2\right]+y^2-6y+13\)
\(=\left(x+1-2y\right)^2+\left(y^2-2y\times3+9\right)+4\)
\(=\left(x+1-2y\right)^2+\left(y-3\right)^2+4\)
Ta có:\(\left(x+1-2y\right)^2\ge0\)với mọi \(x;y\in R\)
và\(\left(y-3\right)^2\ge0\)với mọi \(x;y\in R\)
\(\Rightarrow\left(x+1-2y\right)^2+\left(y-3\right)^2+4\ge4\)với mọi \(x;y\in R\)
\(\Rightarrow x^2+5y^2+2x-4xy-10y+14>0\)
c)
\(5x^2+10y^2-6xy-4x-2y+3=x^2+4x^2+y^2+9y^2-6xy-4x-2y+3\)
\(=\left[\left(2x\right)^2-2\times2x+1\right]+\left(y^2-2y+1\right)+\left[\left(3y\right)^2-2\times3y+x^2\right]+1\)
\(=\left(2x+1\right)^2+\left(y-1\right)^2+\left(3y-x\right)^2+1\)
Ta có \(\left(2x+1\right)^2\ge0\)với mọi \(x\)
\(\left(y-1\right)^2\ge\)với mọi \(y\)
\(\left(3y-x\right)^2\ge0\)với mọi \(x;y\)
và \(1>0\)
\(\Rightarrow5x^2+10y^2-6xy-4x-2y+3>0\)
a. \(x^2+xy+y^2+1=\left(x^2+xy+\frac{1}{4}y^2\right)+\frac{3}{4}y^2+1=\left(x+\frac{1}{4}y\right)^2+\frac{3}{4}y^2+1>0\forall x;y\)(đpcm)
b. \(x^2+5y^2+2x-4xy-10y+14\)
\(=\left[\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1\right]+\left(y^2-6y+9\right)+4\)
\(=\left[\left(x-2y\right)^2-2\left(x-2y\right)+1\right]+\left(y^2-6y+9\right)+4\)
\(=\left(x-2y-1\right)^2+\left(y-3\right)^2+4>0\forall x;y\)(đpcm)
c. tương tự ý b
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
3)
e)
b) Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0
\(1.\)
\(a.\)
\(x^2-2x=x\left(x-2\right)\)
b.
\(3y^3+6xy^2+3x^2y\)
\(=3y\left(y^2+2xy+x^2\right)\)
\(=3y\left(x+y\right)^2\)
\(c.\)
\(x^2-2xy-xy+2y^2\)
\(=x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-y\right)\left(x-2y\right)\)
\(2.\)
\(a.\)
\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(b.\)
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(c.\)
\(x^2-6xy+9y^2-16\)
\(=\left(x^2-6xy+9y^2\right)-4^2\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Tương tự câu \(d,e,g\)
\(3.\)
\(a.\)
\(x^3-2x=0\)
\(\Rightarrow x\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
\(b.\)
\(x\left(x-4\right)+\left(x-4\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(c.\)
\(x\left(x-3\right)+4x-12=0\)
\(\Rightarrow x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Tương tự \(d,e,g\)
1) x2 + 7y2 - 4xy - 2x - 2y + 4 = 0
\(\Leftrightarrow\)[ x2 - 2x.( 2y + 1 ) + 4y2 + 4y +1 ] - 4y2 - 4y - 1 + 7y2 - 2y +4 = 0
\(\Leftrightarrow\) [ x2 - 2x.( 2y +1 ) + ( 2y +1 )2 ] + 3y2 - 6y +3 = 0
\(\Leftrightarrow\) ( x - 2y - 1 )2 + 3.( y2 - 2y + 1 ) = 0
\(\Leftrightarrow\)( x - 2y - 1 )2 + 3.( y - 1 )2 = 0
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-2y-1\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-2y-1=0\\y-1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=2y+1\\y=1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=3\\y=1\end{cases}}\)
Vậy x = 3 , y = 1 thì x2 + 7y2 - 4xy - 2x - 2y + 4 = 0
2) 11x2 + y2 - 6xy - 14x + 2y +9 = 0
\(\Leftrightarrow\)[ y2 - 2y.( 3x - 1 ) + 9x2 - 6x +1 ] + 2x2 - 8x + 8 = 0
\(\Leftrightarrow\)[ y2 - 2y.( 3x - 1 ) + ( 3x - 1 )2 ] + 2.( x2 - 4x + 4 ) = 0
\(\Leftrightarrow\)( y - 3x + 1 )2 + 2.( x - 2 )2 = 0
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(y-3x+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y-3x+1=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y=3x-1\\x=2\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y=5\\x=2\end{cases}}\)
Vậy x = 2 , y = 5 thì 11x2 + y2 - 6xy - 14x + 2y + 9 = 0
f) x2 + 2y2 - 2xy + 2x + 2 - 4y =0
<=>x2 + y2 - 2xy+2x-2y+y2-2y+1+1=0
<=>(x-y)2+2(x-y)+1+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>y=1;x=0
Bạn học thầy Trung phải k nè~~~~
Busted :))))
a, \(x^2+y^2-2x+10y+26=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+10y+25\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+5\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-5\end{cases}}\)
b,\(4x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(2x+y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+y=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+1=0\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=1\end{cases}}\)
c,\(5x^2+9y^2-12xy+4x+4=0\)
\(\Rightarrow\left(x^2+4x+4\right)+\left(4x^2-12xy+9y^2\right)=0\)
\(\Rightarrow\left(x+2\right)^2+\left(2x-3y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\2x-3y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\2.\left(-2\right)-3y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-\frac{4}{3}\end{cases}}\)
d,\(5x^2+9y^2-6xy-4x+1=0\)
\(\Rightarrow\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^x\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(x-3y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2x+1=0\\x-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\-\frac{1}{2}-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{1}{6}\end{cases}}\)