a ) \(x^2-8x+19\)

\(=x^2-2x.4+16+3\)

\(=\left(x-4\right)^2+3\ge3\forall x\left(đpcm\right)\)

b ) \(3x^2-6x+5\)

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

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

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

\(=3\left(x-1\right)^2+2\ge2\forall x\left(đpcm\right)\)

c ) \(x^2+y^2-8x+4y+27\)

\(=\left(x^2-8x+16\right)+\left(y^2+4y+4\right)+7\)

\(=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\forall x\left(đpcm\right)\)

:D

a)\(x^2-8x+19=x^2-2.x.4+16+3=\left(x+4\right)^2+3\)

Vì \(\left(x+4\right)^2\ge0\Rightarrow\left(x+4\right)^2+3\ge3\Rightarrow x^2-8x+19\ge3\)

Vậy x²-8x+19 luôn nhận giá trị dương

mấy câu kia làm tương tự

\(f,F=x^2+9y^2-8x+4y+27\) (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 \(x;y\).

\(Toru\)

bn kham khảo ở đây nha 

Câu hỏi của Mimi - Toán lớp 8 | Học trực tuyến

vào thống kê hoie đáp của mình có chữ màu xanh trng câu hỏi này nhấn zô đó sẽ ra 

hc tốt:~:B~

a) \(x^2-8x+2018=x^2-8x+16+2002=\left(x^2-8x+16\right)+2002=\left(x-4\right)^2+2002\)

Vì \(\left(x-4\right)^2\ge0\)

\(\Rightarrow\left(x-4\right)^2+2002\ge2002\)(Luôn Luôn Dương)

b)\(3x^2+6x+7=3x^2+6x+3+4=3\left(x^2+2x+1\right)+4=3\left(x+1\right)^2+4\)

Vì \(3\left(x+1\right)^2\ge0\)

\(\Rightarrow3\left(x+1\right)^2+4\ge4\)(Luôn Luôn Dương)

c)\(3x^2-6x+5=3x^2-6x+3+2=3\left(x^2-2x+1\right)+2=3\left(x-1\right)^2+2\)

Vì \(3\left(x-1\right)^2\ge0\)

\(\Rightarrow3\left(x-1\right)^2+2\ge2\)(Luôn Luôn Dương)

d)\(x^2-8x+19=x^2-8x+16+3=\left(x^2-8x+16\right)+3=\left(x-4\right)^2+3\)

Vì \(\left(x-4\right)^2\ge0\)

\(\Rightarrow\left(x-4\right)^2+3\ge3\)(Luôn Luôn Dương)

\(a,x^2-8x+2018=\left(x^2-8x+16\right)+2012=\left(x-4\right)^2+2012\)

Vì \(\orbr{\begin{cases}\left(x-4\right)^2\ge0\\2012>0\end{cases}}\Rightarrow\left(x-4\right)^2+2012>0\)

\(\Rightarrow x^2-8x+2018>0\) \(\forall x\)

Ta có : C = 4x² + 4y² - 8x + 4y + 427

=> C = (4x² - 8x + 4) + (4y² + 4y + 1) + 422

=> C = (2x - 2)² + (2y + 1)² + 422

Mà \(\left(2x-2\right)^2\ge0\forall x\)

       \(\left(2y+1\right)^2\ge0\forall x\)

Nên C = (2x - 2)² + (2y + 1)² + 422  \(\ge422\forall x\)

Suy ra : C = (2x - 2)² + (2y + 1)² + 422 \(>0\forall x\)

Vậy C luôn luôn dương (đpcm)

x^2-8x+20=(x^2-8x+16)+4

                 =(x-4)^2+4>0(vì (x-4)^2>=0)

4x^2-12x+11=4x^2-12x+9+2

                     =(2x-3)^2+2>0

x^2-x+1=x^2-x+1/4+3/4

             =(x-1/2)^2+3/4>0

x^2-2x+y^2+4y+6

=x^2-2x+1+y^2+4y+4+1

=(x-1)^2+(y+2)^2+1>0

a: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

\(=\left(x-4\right)^2+4>0\forall x\)

b: Ta có: \(4x^2-12x+11\)

\(=4x^2-12x+9+2\)

\(=\left(2x-3\right)^2+2>0\forall x\)

c: Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

d: Ta có: \(x^2-2x+y^2+4y+6\)

\(=x^2-2x+1+y^2+4y+4+1\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+1>0\forall x,y\)

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

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

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-\left(x^2+2x-8\right)\)

\(=-\left(x^2+2x+1-9\right)\)

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

\(=x^2-2x+1+y^2-4y+4+1\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)

a) \(P=2x-x^2-2\)

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

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

Vì \(-\left(x-1\right)^2\le0;\forall x\)

\(\Rightarrow-\left(x-1\right)^2-1\le0-1;\forall x\)

Hay \(P\le-1< 0;\forall x\)

Vậy biểu thức P luôn có giá trị âm với mọi x

b)  \(Q=-x^2-y^2+8x+4y-21\)

\(=-\left(x^2-8x+16\right)-\left(y^2-4y+4\right)-1\)

\(=-\left(x-4\right)^2-\left(y-2\right)^2-1\)

Vì \(\hept{\begin{cases}-\left(x-4\right)^2\le0;\forall x,y\\-\left(y-2\right)\le0;\forall x,y\end{cases}}\)

\(\Rightarrow-\left(x-4\right)^2-\left(y-2\right)^2\le0;\forall x,y\)

\(\Rightarrow-\left(x-4\right)^2-\left(y-2\right)^2-1\le0-1;\forall x,y\)

Hay \(Q\le-1< 0;\forall x,y\)

Vậy biểu thức Q luôn âm với mọi gt của x,y

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