Ta có : 

\(G=-5x^2+7x-3\)

\(\Rightarrow G=-\left(5x^2+7x+3\right)\)

\(\Rightarrow G=-\left[x^2+2x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\left(\frac{7}{2}\right)^2+4x^2\right]\)

\(\Rightarrow G=-\left[\left(x+\frac{7}{2}\right)^2+\frac{49}{4}-3+4x^2\right]\)

\(\Rightarrow G=-\left[\left(x+\frac{7}{2}\right)^2+\frac{37}{4}+4x^2\right]\)\(\Rightarrow G=-\left(x+\frac{7}{2}\right)^2-\frac{37}{4}-4x^2\)

\(\Rightarrow G< 0\forall x\)

\(H=-4x^2-6x-4\)

\(\Rightarrow H=-\left(4x^2+6x+4\right)\)

\(\Rightarrow H=-\left[\left(2x\right)^2+2.2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{7}{4}\right]\)

\(\Rightarrow H=-\left[\left(2x+\frac{3}{2}\right)^2+\frac{7}{4}\right]\)

\(\Rightarrow H=-\left(2x+\frac{3}{2}\right)^2-\frac{7}{4}< 0\forall x\)

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  \(\Leftrightarrow H>0\)

\(E=-x^2-3x-5=-\left(x^2+3x+5\right)=-\left(x^2+2.\frac{3}{2}x+\frac{9}{4}\right)-\frac{11}{4}\\ \)

\(=-\left(x+\frac{3}{2}\right)^2-\frac{11}{4}=-\left(\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\right)\le-\frac{11}{4}< 0\)

\(F=-3x^2-6x-4=-3.\left(x^2+2x+\frac{4}{3}\right)=-3.\left(\left(x^2+2x+1\right)+\frac{1}{3}\right)\)

\(=-3.\left(\left(x+1\right)^2+\frac{1}{3}\right)\le-\frac{3.1}{3}=-1< 0\)

\(-x^2-3x-5\)

\(=-\left(x^2+3x+5\right)\)

\(=-\left[x^2+2x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+5\right]\)

\(=-\left[\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+5\right]\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{11}{4}\)

Vậy biểu thức luôn âm với mọi giá trị của x.

\(F=-3x^2-6x-4\)

\(F=-\left(3x^2+6x+4\right)\)

\(F=-\left(x^2+6x+9-5+2x^2\right)\)

\(F=-\left[\left(x+3\right)^2+2x^2-5\right]\)

\(F=-\left(x+3\right)^2-2x^2+5\)

Vậy biểu thức luôn âm với mọi giá trị của x.

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: \(=-3\left(x^2+2x+\dfrac{4}{3}\right)=-3\left(x^2+2x+1+\dfrac{1}{3}\right)\)

\(=-3\left(x+1\right)^2-1< 0\)

\(F=2x^2+4x+3\)

\(=2\left(x^2+2x+1\right)+1\)

\(=2\left(x+1\right)^2+1\)\(>\)\(0\)     (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

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.

\(H=4x^2+4x+2=\left(2x+1\right)^2+1>0\)

\(K=4x^2+3x+2=4\left(x^2+2.\frac{3}{8}x+\frac{9}{64}\right)+\frac{23}{16}\)

      \(=4\left(x+\frac{3}{8}\right)^2+\frac{23}{16}>0\)

\(L=2x^2+3x+4=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)+\frac{23}{8}\)

      \(=2\left(x+\frac{3}{4}\right)^2+\frac{23}{8}>0\)

Bạn viết rõ ra câu h đc ko giúp nha thanks bạn nhiều Đườnh Quỳnh Giang