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Ta có: \(M=18+4x-8y+6xy+5x^2+10y^2\)
\(=4x^2+4x+1+x^2+6xy+9y^2+y^2-8y+16+1\)
\(=\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2+1\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\left(x+3y\right)^2\ge0\forall x,y\)
\(\left(y-4\right)^2\ge0\forall y\)
Do đó: \(\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2\ge0\forall x,y\)
\(\Leftrightarrow\left(2x+1\right)^2+\left(x+3y\right)^2+\left(y-4\right)^2+1\ge1>0\forall x,y\)
hay \(M>0\forall x,y\)
\(M=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)
\(M=\left(x+3y\right)^2+\left(2x+1\right)^2+\left(y-4\right)^2+1>0;\forall x;y\)
\(5x^2+10y^2-6xy-4x-10y+14\)
\(=\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^2\right)+\left(y^2-10y+25\right)-12\)
\(=\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-5\right)^2-12\ge-12\) đề có nhầm không bạn?
_______________Bài làm___________________
a, \(x^2+xy+y^2+1\)
\(=\left(x^2+2x\dfrac{y}{2}+\dfrac{y^2}{4}\right)+\dfrac{3y^2}{4}+1=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^3}{4}+1\)
Do \(\left(x+\dfrac{y}{2}\right)^2\ge0\forall x,y\)
Và \(\dfrac{3y^2}{4}\ge0\forall y\)
Nên: \(\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0\forall x,y=>đpcm\)
b, \(x^2+5y^2+2x-4xy-10y+14\)
\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+\left(y^2-6y+9\right)+5\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+\left(y-3\right)^2+5\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4\)
Do \(\left(x-2y+1\right)^2\ge0\forall x,y\)
Và \(\left(y-3\right)^2\ge0\forall y\)
Nên \(\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)
c, \(5x^2+10y^2-6xy-4x-2y+3\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-2x+1\right)+\left(y^2-2y+1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\)
Do .........
tự làm ik
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, \(x^2+xy+y^2+1=x^2+\dfrac{1}{2}xy+\dfrac{1}{2}xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2+1\)
\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\)
Với mọi giá trị của \(x;y\in R\) ta có:
\(\left(x^2+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\)
\(\Rightarrow\left(x^2+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2+1\ge1\)
Vậy............
b, \(5x^2+10y^2-6xy-4x-2y+3\)
\(=x^2-6xy+9y^2+4x^2-4x+1+y^2-2y+1+1\)
\(=x^2-3xy-3xy+9y^2+4x^2-2x-2x+1+y^2-y-y+1+1\)
\(=x\left(x-3y\right)-3y\left(x-3y\right)+2x\left(2x-1\right)-\left(2x-1\right)+y\left(y-1\right)-\left(y-1\right)+1\)
\(=\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\)
Với mọi giá trị của \(x;y\in R\) ta có:
\(\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2\ge0\)
\(\Rightarrow\left(x-3y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
Vậy..............
Chúc bạn học tốt!!!
`#3107.\text {DN}`
a)
\((2x-3)^2-x(3-x)+5x-4x^2+17\)
`= 4x^2 - 12x + 9 - 3x + x^2 + 5x - 4x^2 + 17`
`= x^2 - 10x + 26`
b)
`M = x^2 - 10x + 26`
`= [(x)^2 - 2*x*5 + 5^2] + 1`
`= (x - 5)^2 + 1`
Vì `(x - 5)^2 \ge 0` `AA` `x => (x - 5)^2 + 1 \ge 1` `AA` `x`
Vậy, giá trị biểu thức M luôn có giá trị dương với mọi x.
a, x^2 + xy + y^2 + 1
= (x+y/4) ^2 + 3/4.y^2 + 1 >= 1 > 0
\(M=18+4x-8y+6xy+5x^2+10y^2\)
\(=\left(x^2+6xy+9y^2\right)+\left(4x^2+4x+1\right)+\left(y^2-8y+16\right)+1\)
\(=\left(x+y\right)^2+4\left(x+\frac{1}{2}\right)^2+\left(y-4\right)^2+1\)
Có \(\left(x+y\right)^2\ge0\forall xy\)
\(4\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\left(y-4\right)^2\ge0\forall y\)
\(\Rightarrow M\ge1\forall x,y\)
hay \(M>0\forall x,y\)