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1/
a, \(x^2-6x+10=x^2-6x+9+1=\left(x-3\right)^2+1\ge1>0\)
b,\(4x-x^2-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\)
2/
a, \(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)
Dấu "=" xảy ra khi x-1=0 <=> x=1
Vậy Pmax = 4 khi x = 1
b, \(M=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)^2+\left(y^2+6y+9\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)
Vậy Mmax = 3/4 khi x = 1/2, y = -3
\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)
\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(đpcm\right)\)
\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)
\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)
\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)
a) \(x^2+x+2=\left(x^2+x+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\)đúng \(\forall x\in R\)
b) \(x^2-4x+10=\left(x^2-4x+4\right)+6=\left(x-2\right)^2+6\ge6>0\)đúng \(\forall x\in R\)
c) \(x\left(x-4\right)+10=x^2-4x+10\)(giải như câu b)
d) \(x\left(2-x\right)-4=-\left(x^2-2x+1\right)-3=-\left(x-1\right)^2-3\le-3< 0\)đúng \(\forall x\in R\)
e) \(x^2-5x+2017=\left(x^2-5x+\frac{25}{4}\right)+\frac{8043}{4}=\left(x-\frac{5}{2}\right)^2+\frac{8043}{4}\ge\frac{8043}{4}>0\)đúng \(\forall x\in R\)
a)\(x^2-2xy+y^2+1=\left(x+y\right)^2+1\ge1>0\)
b)\(x-x^2-1=-\left(x^2-x+\frac{1}{4}\right)^2-\frac{3}{4}\le-\frac{3}{4}< 0\)
c)\(9x^2+12x+10=\left(9x^2+12x+4\right)+6=\left(3x+2\right)^2+6\ge6>0\)
d)\(3x^2-x+1=2x^2+\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=2x^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0`\)
a)
Đặt \(A=9x^2-6x+2\)
\(=\left(3x\right)^2-2.3x+1+1\)
\(=\left(3x+1\right)^2+1\)
Ta có: \(\left(3x+1\right)^2\ge0;\forall x\)
\(\Rightarrow\left(3x+1\right)^2+1\ge0+1;\forall x\)
Hay \(A\ge1>0;\forall x\)
Các phần khác tương tự cứ việc biến đổi thành hằng đẳng thức
\(a,9x^2-6x+2\)
\(=\left(3x\right)^2-2.3x.1+1^2+1\)
\(=\left(3x-1\right)^2+1\)
Vì\(\left(3x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(3x-1\right)^2+1\ge1>0\forall x\)
\(\Rightarrow9x^2-6x+2>0\forall x\)
\(b,x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
\(\Rightarrow x^2+x+1>0\forall x\)
\(B=4x^2+y^2+12x-4xy-6y+16\)
\(=\left(4x^2+y^2+9-4xy-6y+12x\right)+7\)
\(=\left[\left(2x\right)^2+y^2+3^2-2.2x.y-2.y.3+2.2x.3\right]+7\)
\(=\left(2x-y+3\right)^2+7\)
Ta có :
\(\left(2x-y+3\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(2x-y+3\right)^2+7\ge7>0\forall x,y\)
Hay B > 0 với mọi x,y
Ta có : \(B=\left(2x\right)^2-2.2x\left(y-3\right)+\left(y-3\right)^2-\left(y-3\right)^2+y^2-6y+16\)
\(=\left(2x-y+3\right)^2-y^2+6y-9+y^2-6y+16\)
\(=\left(2x-y+3\right)^2+7\)
Vì \(\left(2x-y+3\right)^2\ge0\forall x,y\Rightarrow B\ge7\)
hay B > 0 với mọi x,y
\(A=9x^2-6x+2=\left(3x\right)^2-2.3x+1+1=\left(3x-1\right)^2+1>0\forall x\)
Vậy ta có đpcm
\(B=x^2-2xy+y^2+1=\left(x-y\right)^2+1>0\forall x;y\)
Vậy ta có đpcm
1) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow 4x^2 + 14x - 10x - 35=4x^2-25\)
\(\Leftrightarrow4x^2-4x^2+14x-10x=35-25\)
\(\Leftrightarrow4x=10\)
\(\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
2) \(x^2-4x+5\)
\(=-(4x-x^2-5 )\)
\(= -[-(x^2-4x)-5 ]\)
\(=-[ -(x^2-2x.2+4-4)-5 ]\)
\(= -[-(x-2)^2+4-5 ]\)
\(= -[-(x-2)^2-1 ]\)
Vì \(-(x-2)^2 ≤0\)\(\forall x\) \(\Rightarrow\) \(-(x-2)^2-1<0\) \(\forall x\)
\(\Rightarrow\)\(-[-(x-2)^2-1 ]>0\)\(\forall x\)
\(\Rightarrow x^2-4x+5>0\)\(\forall x\)
a) Ta có:
\(x^2+4x+5\)
\(=x^2+2.x.2+4+1\)
\(=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+1>0\forall x\)
\(\Rightarrow x^2+4x+5>0\forall x\)
b) Ta có:
\(x^2-x+1\)
\(=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
\(\Rightarrow x^2-x+1>0\forall x\)
c) Ta có:
\(12x-4x^2-10\)
\(=-\left(4x^2-12x+10\right)\)
\(=-\left[\left(2x\right)^2-2.2x.3+9+1\right]\)
\(=-\left(2x-3\right)^2-1\)
Vì \(-\left(2x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(2x-3\right)^2-1< 0\forall x\)
\(\Rightarrow12x-4x^2-10< -1\)