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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
\(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1>0\forall x;y\)
\(x^2+5y^2+2x-4xy-10y+14\)
\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+y^2-6y+9+4\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\forall x;y\)
Chúc bạn học tốt.
Tham khảo bài làm của mình : Câu hỏi của Phạm Bá Gia Nhất - Toán lớp 8 - Học toán với OnlineMath
\(A=4x^2+4x+11\)
\(=\left(4x^2+4x+1\right)+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Min A = 10 khi: 2x + 1 = 0
<=> x = -1/2
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,\(5x^2+9y^2-6xy-4x+1=0\)
\(\Rightarrow\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^x\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(x-3y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2x+1=0\\x-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\-\frac{1}{2}-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{1}{6}\end{cases}}\)
a) 5x2 + 10y2 - 6xy - 4x - 2y + 3
= ( x2 - 6xy + 9y2 ) + ( 4x2 - 4x + 1 ) + ( y2 - 2y + 1 ) + 1
= ( x - 3y )2 + ( 2x - 1 )2 + ( y - 1 )2 + 1
Ta có : \(\hept{\begin{cases}\left(x-3y\right)^2\\\left(2x-1\right)^2\\\left(y-1\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1>0\forall x,y\)
=> đpcm
b) x2 + 4y2 + z2 - 2x - 6z + 8y + 15 = 0 < Sửa -z2 -> +z2 )
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + 4( y2 + 2y + 1 ) + ( z - 3 )2 + 1
= ( x - 1 )2 + 4( y + 1 )2 + ( z - 3 )2 + 1
Ta có : \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\4\left(y+1\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\Rightarrow\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge1>0\forall x,y,z\)
=> đpcm
\(5x^2+10y^2-6xy-4x-10y+14\)
\(=\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^2\right)+\left(y^2-10y+25\right)-12\)
\(=\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-5\right)^2-12\ge-12\) đề có nhầm không bạn?
a) 5x2 + 10y2 - 6xy - 4x - 2y + 3
= ( x2 - 6xy + 9y2 ) + ( 4x2 - 4x + 1 ) + ( y2 - 2y + 1 ) + 1
= ( x - 3y )2 + ( 2x - 1 )2 + ( y - 1 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm
b) x2 + 4y2 + z2 - 2x - 6z + 8y + 15
= ( x2 - 2x + 1 ) + ( 4y2 + 8y + 4 ) + ( z2 - 6z + 9 ) + 1
= ( x - 1 )2 + ( 2y + 2 )2 + ( z - 3 )2 + 1 ≥ 1 > 0 ∀ x, y, z
=> đpcm