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a. \(2x^2-4x+10=x^2-2x+1+x^2-2x+1+8=\left(x-1\right)^2+\left(x-1\right)^2+8=2\left(x-1\right)^2+8\)
Vì \(2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2+8\ge8\)
Vậy...
b. \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy..
c. \(2x^2-6x+5=x^2-4x+4+x^2-2x+1=\left(x-2\right)^2+\left(x-1\right)^2\)
Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}}\Rightarrow\left(x-2\right)^2+\left(x-1\right)^2\ge0\)
Vậy...
a) Đặt \(A=x^2+4x+7\)
\(A=\left(x^2+4x+4\right)+3\)
\(A=\left(x+2\right)^2+3\)
Mà \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow A\ge3>0\)
b) Đặt \(B=4x^2-4x+5\)
\(B=\left(4x^2-4x+1\right)+4\)
\(B=\left(2x-1\right)^2+4\)
Mà \(\left(2x-1\right)^2\ge0\forall x\)
\(\Rightarrow B\ge4>0\)
c) Đặt \(C=x^2+2y^2+2xy-2y+3\)
\(C=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x+y\right)^2+\left(y-1\right)^2+2\)
Mà \(\left(x+y\right)^2\ge0\forall x;y\)
\(\left(y-1\right)^2\ge0\forall y\)
\(\Rightarrow C\ge2>0\)
\(A=x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\)
Vậy \(A_{min}=1\Leftrightarrow x=-1\)
\(B=x^2+4x=6=x^2+4x+4+2=\left(x+2\right)^2+2\ge2>0\)
Vậy \(B_{min}=2\Leftrightarrow x=-2\)
\(a;x^2-3x+3=x^2-2\cdot\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+3\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\Leftrightarrow x^2-3x+3>0\forall x\)
\(a,-x^2+6x-16\)
\(=-x^2+3x+3x-9-5\)
\(=-x\left(x-3\right)+3\left(x-3\right)-5\)
\(=\left(3-x\right)\left(x-3\right)-5\)
\(=-\left(x-3\right)^2-5\le-5\)=>Luôn âm
\(c,-1+x-x^2\)
\(=-x^2+x-1\)
\(=-\left(x^2-x+\frac{1}{2}+\frac{1}{2}\right)\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\le\frac{-1}{2}\)=>Luôn âm
Cm: Ta có:
a) A = x2 - 8x + 20 = (x2 - 8x + 16) + 4 = (x - 4)2 + 4 > 0 \(\forall\) x(vì (x - 4)2 \(\ge\)0 \(\forall\)x ; 4 > 0)
=> A luôn dương với mọi x
b) B = 4x2 - 12x + 11 = [(2x)2 - 12x + 9] + 2 = (2x - 3)2 + 2 > 0 \(\forall\)x (vì (2x - 3)2 \(\ge\)0 \(\forall\)x; 2 > 0)
=> B luôn dương với mọi x
c) C = x2 - x + 1 = (x2 - x + 1/4) + 3/4 = (x - 1/2)2 + 3/4 > 0 \(\forall\)x (vì (x - 1/2)2 \(\ge\)0 \(\forall\)x; 3/4 > 0)
=> C luôn dương với mọi x
* Tìm x
3(x + 2)2 + (2x - 1)2 - 7(x + 3)(x - 3) = 36
=> 3(x2 + 4x + 4) + 4x2 - 4x + 1 - 7(x2 - 9) = 36
=> 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63 = 36
=> 8x + 76 = 36
=> 8x = 36 - 76
=> 8x = -40
=> x = -40 : 8 = -5
a) \(2x^2-4x+10=2\left(x^2-2x+1\right)+8=2\left(x-1\right)^2+8>0\)
b) \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
c) \(2x^2-6x+5=2\left(x^2-3x+\dfrac{9}{4}\right)+1,5=2\left(x-\dfrac{3}{2}\right)^2+1,5>0\)
a. \(x^2-8x+19\)
\(=x^2-2.x.4+16+3\)
\(=\left(x-4\right)^2+3\ge3\forall x\)
=> đpcm
b. \(4x^2+4x+3\)
\(=\left(2x\right)^2+2.2x.1+1+2\)
\(=\left(2x+1\right)^2+2\ge2\forall x\)
=> đpcm
\(A=-x^2+4x+11\)
\(-A=x^2-4x-11\)
\(-A=\left(x^2-4x+4\right)-15\)
\(-A=\left(x-2\right)^2-15\)
Mà \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow-A\ge-15\Leftrightarrow A\le15\)
Vậy ...( kiểm tra lại đề -__- )
\(B=5x-x^2-10\)
\(-B=x^2-5x+10\)
\(-B=\left(x^2-5x+\frac{25}{4}\right)+\frac{15}{4}\)
\(-B=\left(x-\frac{5}{2}\right)^2+\frac{15}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow-B\ge\frac{15}{4}\Leftrightarrow B\le-\frac{15}{4}< 0\)
Vậy ...
\(A=-x^2+4x+11=-\left(x^2-4x-11\right)=-\left(x^2-4x+4\right)+15\)
\(-\left(x-2\right)^2+15=15-\left(x-2\right)^2\)
\(Mà\left(x-2\right)^2\ge0\left(\forall x\right)\Rightarrow15-\left(x-2\right)^2\le15\)
(Đề có vấn đề tí)
\(B=5x-x^2-10=-\left(x^2-5x+10\right)=-\left(x-\frac{5}{2}\right)^2-\frac{15}{4}\)
Vì \(\left(x-\frac{5}{2}\right)^2\ge0\left(\forall x\right)\Rightarrow-\left(x-\frac{5}{2}\right)^2\le0\Rightarrow-\left(x-\frac{5}{2}\right)^2-\frac{15}{4}< 0\)
Vậy biểu thức trên không dương với mọi x