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\(P=16x^2+8x+2=\left(16x^2+8x+1\right)+1=\left(4x+1\right)^2+1\)
Do \(\left\{{}\begin{matrix}\left(4x+1\right)^2\ge0\\1>0\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow P=\left(4x+1\right)^2+1>0;\forall x\) (đpcm)
C=(4x)2+4x+1+99
=(4x+1)2+99>0
Vậy biểu thức luôn dương
Chúc hok tốt
Xét \(C=16x^2+4x+100\)
\(C=4x\left(4x+1\right)+100\)
Mà \(4x\left(4x+1\right)\ge0,\forall x\)( \(\forall x\)nghĩa là VỚI MỌI X nha bạn)
\(\Rightarrow4x\left(4x+1\right)+100>0,\forall x\)
\(\Leftrightarrow C>0\)
Vậy, \(4x\left(4x+1\right)+100>0,\forall x\)(ĐPCM)
\(f,F=x^2+9y^2-8x+4y+27\) (sửa đề)
\(=\left(x^2-8x+16\right)+\left(9y^2+4y+\dfrac{4}{9}\right)+\dfrac{95}{9}\)
\(=\left(x^2-2\cdot x\cdot4+4^2\right)+\left[\left(3y\right)^2+2\cdot3y\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2\right]+\dfrac{95}{9}\)
\(=\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2+\dfrac{95}{9}\)
Ta thấy: \(\left(x-4\right)^2\ge0\forall x\)
\(\left(3y+\dfrac{2}{3}\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2+\dfrac{95}{9}\ge\dfrac{95}{9}>0\forall x;y\)
hay \(F\) luôn dương với mọi \(x;y\).
\(Toru\)
Ta có :
\(B=x^2-10x+28\)
\(\Rightarrow B=x^2-2.x.5+25+3\)
\(\Rightarrow B=\left(x+5\right)^2+3\)
Vì \(\left(x+5\right)\ge0\) ( với mọi x )
\(\Rightarrow\left(x+5\right)+3\ge3\)
=> đpcm
\(A=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)
\(B=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)
\(C=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)
\(D=\left(x-5\right)^2+\left(3y+1\right)^2+4\)
\(E=\left(4x+1\right)^2+\left(y-2\right)^2+1\)
\(M=-\left(x+\frac{7}{2}\right)^2-\frac{11}{4}\)
\(N=-5\left(x-\frac{3}{5}\right)^2-\frac{41}{5}\)
\(C\) đề sai ví dụ \(x=3\Rightarrow C=2>0\)
\(D=-5\left(x-\frac{7}{10}\right)^2-\frac{131}{20}\)
a)\(A=x^2+x+1=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
b) \(B=2x^2+2x+1=2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{2}=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\)
\(2,B=x^2-10x+27\)
\(=x^2-2.x.5+5^2+2\)
\(=\left(x-5\right)^2+2\)
Ta thấy: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-5\right)^2+2\ge2\forall x\)
hay B luôn dương
\(4,D=-16x^2+16x-9\)
\(=-\left[\left(4x\right)^2-2.4x.2+2^2\right]-5\)
\(=-\left(4x-2\right)^2-5\)
Ta thấy: \(\left(4x-2\right)^2\ge0\forall x\)
\(\Rightarrow-\left(4x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(4x-2\right)^2-5\le-5\forall x\)
hay D luôn âm.
2: B=x^2-10x+25+2
=(x-5)^2+2>=2>0 với mọi x
=>B luôn dương với mọi x
4: D=-16x^2+16x-9
=-(16x^2-16x+9)
=-(16x^2-16x+4+5)
=-(4x-2)^2-5<=-5<0
=>D luôn âm với mọi x