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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) \(2x^2-4x+10=2\left(x^2-2x+1\right)+8=2\left(x-1\right)^2+8>0\)
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,\(-\left(x^2-3x+4\right)\)
\(-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\Leftrightarrow-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)(luôn âm)
b\(-2\left(x^2-5x+\frac{15}{2}\right)\)
\(-2\left[\left(x-\frac{5}{2}\right)^2+\frac{5}{4}\right]\)
\(-2\left(x-\frac{5}{4}\right)^2-\frac{5}{2}\le-\frac{5}{2}\)(luôn âm)
c,\(-\left[\left(4x^2-4x+1\right)+\left(2y^2-6y+5\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y^2-3y+\frac{5}{2}\right)\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2+\frac{1}{4}\right]\)
\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2\right]-\frac{1}{4}\le-\frac{1}{4}\)(luôn âm)
\(a,-x^2+6x-16\)
\(=-x^2+3x+3x-9-5\)
\(=-x\left(x-3\right)+3\left(x-3\right)-5\)
\(=\left(3-x\right)\left(x-3\right)-5\)
\(=-\left(x-3\right)^2-5\le-5\)=>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
a : x2 + 4x + 7 = (x + 2)2 + 3 > 0
b : 4x2 - 4x + 5 = (2x - 1)2 + 4 > 0
c : x2 + 2y2 + 2xy - 2y + 3 = (x + y)2 + (y - 1)2 + 2 > 0
d : 2x2 - 4x + 10 = 2(x - 1)2 + 8 > 0
e : x2 + x + 1 = (x + 0,5)2 + 0,75 > 0
f : 2x2 - 6x + 5 = 2(x - 1,5)2 + 0,5 > 0
2. Ta có: P = 2x2 + y2 - 4x - 4y + 10
P = 2(x2 - 2x + 1) + (y2 - 4y + 4) + 4
P = 2(x - 1)2 + (y - 2)2 + 4 \(\ge\)4 \(\forall\)x;y
=> P luôn dương với mọi biến x;y
3 Ta có:
(2n + 1)(n2 - 3n - 1) - 2n3 + 1
= 2n3 - 6n2 - 2n + n2 - 3n - 1 - 2n3 + 1
= -5n2 - 5n = -5n(n + 1) \(⋮\)5 \(\forall\)n \(\in\)Z
Ta có : 9x2 - 6x + 5
= (3x)2 - 6x + 1 + 4
= (3x - 1)2 + 4
Mà : (3x - 1)2 \(\ge0\forall x\)
Nên : (3x - 1)2 + 4 \(\ge4\forall x\)
Suy ra : (3x - 1)2 + 4 \(>0\forall x\)
Vậy biểu thức sau luôn luôn dương
a) 9x2 - 6x + 2 = (3x)2 - 2.3x.1 + 12 + 1 = (3x - 1)2 + 1 mà\(\left(3x+1\right)^2\ge0\Rightarrow\left(3x+1\right)^2+1\ge1>0\)
b) x2 + x + 1 = x2 + 2.x.\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)mà\(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
c) 2x2 + 2x + 1 =\(\left(\sqrt{2}x\right)^2+2\sqrt{2}x.\frac{1}{\sqrt{2}}+\left(\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\)
a) \(9x^2-6x+2=\left(\left(3x\right)^2-2.3x.1+1\right)+1=\left(3x-1\right)^2+1>0\)
b) .\(x^2+x+1=\left(\left(x^2\right)+2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
c) \(2x^2+2x+1=x^2+\left(x^2+2x+1\right)=x^2+\left(x+1\right)^2>0\)
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...