C = -3x² - 6x - 12

    = -3( x² + 2x + 1 ) - 9

    = -3( x + 1 )² - 9 ≤ -9 < 0 ∀ x ( đpcm )

D = -4x² - 12x - 15

     = -4( x² + 3x + 9/4 ) - 6

     = -4( x + 3/2 )² - 6 ≤ -6 < 0 ∀ x ( đpcm )

E = -30 - 5x² + 10x

    = -5( x² - 2x + 1 ) - 25

    = -5( x - 1 )² - 25 ≤ -25 < 0 ∀ x ( đpcm )

\(C=-3x^2-6x-12\)

\(\Rightarrow C=-\left(3x^2+6x+12\right)\)

\(\Rightarrow C=-\left(3x^2+6x+3+9\right)\)

\(\Rightarrow C=-\left[3\left(x+1\right)^2+9\right]\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow3\left(x+1\right)^2+9\ge9\)

\(\Rightarrow C=-\left[3\left(x+1\right)^2+9\right]\le-9\)

=> Đpcm

\(D=-4x^2-12x-15\)

\(\Rightarrow D=-\left(4x^2+12x+15\right)\)

\(\Rightarrow D=-\left[4\left(x+\frac{3}{2}\right)^2+6\right]\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow4\left(x+\frac{3}{2}\right)^2+6\ge6\)

\(\Rightarrow D=-\left[4\left(x+\frac{3}{2}\right)^2+6\right]\le-6\)

=> Đpcm

\(E=-30-5x^2+10x\)

\(\Rightarrow E=-\left(5x^2-10x+30\right)\)

\(\Rightarrow E=-\left[5\left(x-1\right)^2+25\right]\)

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow5\left(x-1\right)^2+25\ge25\)

\(\Rightarrow E=-\left[5\left(x-1\right)^2+25\right]\le-25\)

=> Đpcm

1/

\(M=3x^2-4x+3=3\left(x^2-\frac{4}{3}x+1\right)=3\left(x^2-2x\cdot\frac{2}{3}+\frac{4}{9}\right)+\frac{5}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)

\(N=5x^2-10x+2018=5\left(x^2-2x+1\right)+2013=5\left(x-1\right)^2+2013\ge2013>0\)

\(P=x^2+2y^2-2xy+4y+7=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+3=\left(x-y\right)^2+\left(y+2\right)^2+3\ge3>0\)

2/

\(A=10x-6x^2+7=-6x^2+10x+7=-6\left(x^2-\frac{10}{6}x+\frac{25}{36}\right)-\frac{11}{6}=-6\left(x-\frac{5}{6}\right)^2-\frac{11}{6}\le-\frac{11}{6}< 0\)

\(B=-3x^2+7x+10=-3\left(x^2-\frac{7}{3}x+\frac{49}{36}\right)-\frac{311}{12}=-3\left(x-\frac{7}{6}\right)^2-\frac{311}{12}\le-\frac{311}{12}< 0\)

\(C=2x-2x^2-y^2+2xy-5=\left(2x-x^2-1\right)-\left(x^2-2xy+y^2\right)-4=-\left(x^2-2x+1\right)-\left(x-y\right)^2-4=-\left(x-1\right)^2-\left(x-y\right)^2-4\)\(\le-4< 0\)

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

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

vậy bt A luôn......

\(3x-x^2-7=-\left(x^2-3x+7\right)=-\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)-\frac{19}{4}\)

\(=-\left(x-\frac{3}{2}\right)^2-\frac{19}{4}\le-\frac{19}{4}\)với mọi \(x\).

Do đó ta có đpcm.

Ta có E = \(3x^2+x+5=3\left(x^2+\frac{x}{3}+\frac{5}{3}\right)=3\left(x^2+2.x.\frac{1}{6}+\frac{1}{36}+\frac{59}{36}\right)\)

\(=3\left(x+\frac{1}{6}\right)^2+\frac{59}{12}\ge\frac{59}{12}>0\)

=> E luôn dương với mọi x 

Trả lời:

\(E=3x^2+x+5=3\left(x^2+\frac{1}{3}x+\frac{5}{3}\right)=3\left(x^2+2.x.\frac{1}{6}+\frac{1}{36}+\frac{59}{36}\right)\)

\(=3\left[\left(x+\frac{1}{6}\right)^2+\frac{59}{36}\right]=3\left(x+\frac{1}{6}\right)^2+\frac{59}{12}\)

Ta có: \(\left(x+\frac{1}{6}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(x+\frac{1}{6}\right)^2\ge0\forall x\)

\(\Rightarrow3\left(x+\frac{1}{6}\right)^2+\frac{59}{12}\ge\frac{59}{12}>0\forall x\)

Dấu "=" xảy ra khi x + 1/6 = 0 <=> x = - 1/6

Vậy biểu thức E luôn dương.

\(a;x^2-3x+3=x^2-2\cdot\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+3\)

                 \(=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\Leftrightarrow x^2-3x+3>0\forall x\)

a, TA CO X ² -3X+3=X²-3X+(3/2)² +3/4=(X-3/2)²+3/4 >0

TUONG TU

Lần sau bạn ra ít một thôi nha.Mik chia ra làm 2 nha !

b

\(x^2+5x+7\)

\(=\left(x^2+2\cdot\frac{5}{2}x+\frac{25}{4}\right)+\frac{3}{4}\)

\(=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}>0\)

c

\(x^2-x+\frac{1}{3}\)

\(=\left(x^2-2\cdot\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{12}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{1}{12}>0\)

d

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

\(=3\left(x^2+2\cdot\frac{5}{6}\cdot x+\frac{25}{36}\right)-\frac{3}{36}\)

\(=3\left(x+\frac{5}{6}\right)^2-\frac{3}{36}\left(trueorfalse??\right)\)

e

\(4x^2+x+1\)

\(=4\left(x^2+2\cdot\frac{1}{8}\cdot x+\frac{1}{64}\right)+\frac{15}{16}\)

\(=4\left(x+\frac{1}{8}\right)^2+\frac{15}{16}>0\)

Câu d đề sai nha !

A=(x+2)^2 +3

B=(x-5)^2 +4

C=4(x+1/2)^2 +4

D=(x-1/2)^2 +19/4

E=2(x-3/4)^2 +95/8

Bài 1

\(a,\)\(49x^2-28x+7\)

\(=\left(7x\right)^2-2.7x.2+2^2+3\)

\(=\left(7x-2\right)^2+3\ge3\)( luôn dương )

Dấu bằng sảy ra khi và chỉ khi \(\left(7x-2\right)^2=0\)

\(\Rightarrow7x-2=0\)

\(\Rightarrow x=\frac{2}{7}\)

Bài 1 b

\(x^2+\frac{2}{5}x+\frac{1}{5}\)

\(=x^2+2.x.\frac{1}{5}+\frac{1}{25}+\frac{4}{25}\)

\(=\left(x+\frac{1}{5}\right)^2+\frac{4}{25}\ge\frac{4}{25}\)( luôn dương )

Dấu bằng sảy ra khi và chỉ khi \(\left(x+\frac{1}{5}\right)^2=0\)

\(\Rightarrow x+\frac{1}{5}=0\)

\(\Rightarrow x=-\frac{1}{5}\)

Bài 2 a

\(-9x^2+24x-12\)

\(=-\left(3x^2-2.3x.4+4^2-4\right)\)

\(-\left[\left(3x-4\right)^2-4\right]\)

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

Sai đề chăng ?