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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
1a) 8xy(8-12x+6x*x-x*x*x)
chú thích x*x là x bình phương
x*x*x là x lập phương
2. a) 3x (x-5)- (x-1)(2+3x)=30
3x*x-15x-2x-3x*x+2+3x=30
14x=28
x=2
b) (x+2)(x-3)-(x-2)(x+5)=0
x*x-3x+2x-6-x*x-5x+2x+10=0
2x=-4
x=-2
còn mấy bài còn lại mình không biết
a) x2-6x+10
=(x^2-6x+9)+1
=(x-3)^2+1
vì (x-3)^2>=0 với mọi x nên (x-3)^2+1>0
Hay x^2-6x+10>0
Mình biết hơi muộn
\(A=x^2+2xy+6x+6y+2y^2+8\Leftrightarrow x^2+2xy+6x+6y+y^2+9-1\)
\(A=0\Rightarrow\left(x+y+3\right)^2+y^2-1=0\)
\(\Rightarrow-1\le x+y+3\le1\) .
\(\Rightarrow2012\le x+y+3+2013\le2014\)
\(\Rightarrow2012\le B\le2014\)
\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)
\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)
\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)
ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)
Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)
T i c k cho mình 1 cái nha mới bị trừ 50 đ
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
a) x2 - 2xy + y2 + 1 = (x-y)2 + 1 \(\ge\)1
=> (x-y)2 +1 >0 => x2 - 2xy + y2 >0
b) x - x2 - 1 = -(x2 - x + \(\frac{1}{4}\)) - \(\frac{3}{4}\)= - (x-\(\frac{1}{2}\))2 - \(\frac{3}{4}\)< 0 => x - x2 - 1 <0
a) Ta có:
\(x^2-2xy+y^2+1\)
\(=\left(x^2-2xy+y^2\right)+1\)
.\(=\left(x-y\right)^2+1\)
\(\left(x-y\right)^2\ge0\)với mọi \(x,y\in R\)
\(\Rightarrow x^2-2xy+y^2+1\)
\(=\left(x-y\right)^2+1\ge0+1=1>0 \forall x,y\in R\left(đpcm\right)\)
b) Ta có :
\(x-x^2-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{2^2}+1-\frac{1}{2^2}\right)\)
\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)
Ta có :
\(\left(x-\frac{1}{2}\right)^2\ge0\)với mọi số thực x
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}>0\)với mọi số thực x
\(\Rightarrow x-x^2-1=-\left[\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\right]< 0\)với mọi số thực ( đpcm )
a/ \(x+4y=1\Rightarrow x=1-4y\)
\(A=x^2+4y^2=\left(1-4y\right)^2+4y^2=20y^2-8y+1\)
\(A=20\left(y^2-2.\frac{1}{5}y+\frac{1}{25}\right)+\frac{1}{5}=20\left(y-\frac{1}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}\)
\(\Rightarrow A_{min}=\frac{1}{5}\) khi \(\left\{{}\begin{matrix}y=\frac{1}{5}\\x=1-4y=\frac{1}{5}\end{matrix}\right.\)
b/
\(B=\frac{2x^2+5x+8}{x}=2x+\frac{8}{x}+5\ge2\sqrt{2x.\frac{8}{x}}+5=13\)
\(\Rightarrow B_{min}=13\) khi \(x=2\)