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1: \(=-\left(x^2+2x+2\right)\)
\(=-\left(x^2+2x+1+1\right)\)
\(=-\left(x+1\right)^2-1< 0\)
2: \(=-\left(x^2+4x+7\right)\)
\(=-\left(x^2+4x+4+3\right)\)
\(=-\left(x+2\right)^2-3< 0\)
3: \(=-\left(x^2+6x+11\right)\)
\(=-\left(x^2+6x+9+2\right)\)
\(=-\left(x+3\right)^2-2< 0\)
H========(2x)2+2.2x+1+12+1=(2x+1)2+1>0 hay H luôn là số dương.
\(H=4x^2+4x+2\)
\(H=4x^2+4x+1+1\)
\(H=\left(2x+1\right)^2+1\)
Vậy biểu thức trên luôn dương với mọi x.
\(H=4x^2+4x+2=\left(2x+1\right)^2+1>0\)
\(K=4x^2+3x+2=4\left(x^2+2.\frac{3}{8}x+\frac{9}{64}\right)+\frac{23}{16}\)
\(=4\left(x+\frac{3}{8}\right)^2+\frac{23}{16}>0\)
\(L=2x^2+3x+4=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)+\frac{23}{8}\)
\(=2\left(x+\frac{3}{4}\right)^2+\frac{23}{8}>0\)
5: \(=-\left(x^2+3x+5\right)\)
\(=-\left(x^2+3x+\dfrac{9}{4}+\dfrac{11}{4}\right)\)
\(=-\left(x+\dfrac{3}{2}\right)^2-\dfrac{11}{4}< 0\)
6: \(=-3\left(x^2+2x+\dfrac{4}{3}\right)=-3\left(x^2+2x+1+\dfrac{1}{3}\right)\)
\(=-3\left(x+1\right)^2-1< 0\)
Ta có :
\(G=-5x^2+7x-3\)
\(\Rightarrow G=-\left(5x^2+7x+3\right)\)
\(\Rightarrow G=-\left[x^2+2x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\left(\frac{7}{2}\right)^2+4x^2\right]\)
\(\Rightarrow G=-\left[\left(x+\frac{7}{2}\right)^2+\frac{49}{4}-3+4x^2\right]\)
\(\Rightarrow G=-\left[\left(x+\frac{7}{2}\right)^2+\frac{37}{4}+4x^2\right]\)\(\Rightarrow G=-\left(x+\frac{7}{2}\right)^2-\frac{37}{4}-4x^2\)
\(\Rightarrow G< 0\forall x\)
\(H=-4x^2-6x-4\)
\(\Rightarrow H=-\left(4x^2+6x+4\right)\)
\(\Rightarrow H=-\left[\left(2x\right)^2+2.2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\Rightarrow H=-\left[\left(2x+\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\Rightarrow H=-\left(2x+\frac{3}{2}\right)^2-\frac{7}{4}< 0\forall x\)
\(F=2x^2+4x+3\)
\(=2\left(x^2+2x+1\right)+1\)
\(=2\left(x+1\right)^2+1\)\(>\)\(0\) (với mọi x)
\(G=3x^2-5x+3\)
\(=3\left(x^2-\frac{5}{3}x\right)+3\)
\(=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)+\frac{11}{12}\)
\(=3\left(x-\frac{5}{6}\right)^2>0\) với mọi x
\(F=2x^2+4x+3\)
\(=2\left(x^2+2x+\frac{3}{2}\right)\)
\(=2\left(x+1\right)^2+1\ge1>0\)
vay F luon duong voi moi gt cua x
\(G=3x^2-5x+3=3\left(x^2-\frac{5}{3}x+1\right)=3\left(x^2-2x\frac{5}{6}+\frac{25}{36}+\frac{11}{36}\right)\)
\(=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\ge\frac{11}{12}>0\)
vay......................................
neu co sai bn thong cam nha
CMR các bt sau có gtri âm với mọi gtri của x
5, E=\(-x^2-3x-5\)
6, F=\(-3x^2-6x-4\)
7, G=\(-5x^2+7x-3\)
\(E=-x^2-3x-5=-\left(x^2+3x+5\right)=-\left(x^2+2.\frac{3}{2}x+\frac{9}{4}\right)-\frac{11}{4}\\ \)
\(=-\left(x+\frac{3}{2}\right)^2-\frac{11}{4}=-\left(\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\right)\le-\frac{11}{4}< 0\)
\(F=-3x^2-6x-4=-3.\left(x^2+2x+\frac{4}{3}\right)=-3.\left(\left(x^2+2x+1\right)+\frac{1}{3}\right)\)
\(=-3.\left(\left(x+1\right)^2+\frac{1}{3}\right)\le-\frac{3.1}{3}=-1< 0\)
\(-x^2-3x-5\)
\(=-\left(x^2+3x+5\right)\)
\(=-\left[x^2+2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+5\right]\)
\(=-\left[\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+5\right]\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{11}{4}\)
Vậy biểu thức luôn âm với mọi giá trị của x.
7 )
Ta có :
\(G=-5x^2+7x-3\)
\(\Rightarrow G=-\left(5x^2+7x+3\right)\)
\(\Rightarrow G=-\left[x^2+2x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\left(\frac{7}{2}\right)^2+4x^2\right]\)
\(\Rightarrow G=-\left[\left(x+\frac{7}{2}\right)^2+\frac{49}{4}-3+4x^2\right]\)
\(\Rightarrow G=-\left[\left(x+\frac{7}{2}\right)^2+\frac{37}{4}+4x^2\right]\)\(\Rightarrow G=-\left(x+\frac{7}{2}\right)^2-\frac{37}{4}-4x^2\)
\(\Rightarrow G< 0\forall x\)
8 )
Đề sai nhé bạn :
Nếu thay \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)vào H \(\Leftrightarrow H>0\)
1) câu này sai đề hả bn? -.-
\(2)B=-x^2-4x-7\)
\(B=-\left(x^2+4x+7\right)\)
\(B=-\left(x^2+4x+4+3\right)\)
\(B=-\left[\left(x+2\right)^2+3\right]\)
\(B=-\left(x+2\right)^2-3\)
Vậy biểu thức trên luôn âm với mọi giá trị của x.
\(3)C=-x^2-6x-11\)
\(C=-\left(x^2+6x+11\right)\)
\(C=-\left(x^2+6x+9+2\right)\)
\(C=-\left[\left(x+3\right)^2+2\right]\)
\(C=-\left(x+3\right)^2-2\)
Vậy biểu thức trên luôn âm với mọi x.