-3x^2+2x-1

=-3(x^2-2/3x+1/3)

=-3(x^2-2*x*1/3+1/9+2/9)

=-3(x-1/3)^2-2/3<=-2/3<0 với mọi x

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

\(=\left(x^2+x\right)^2-4\left(x^2+x\right)+4=\left(x^2+x-2\right)^2\ge0\)

\(\Rightarrow\)ĐPCM

Bài 1

\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)

\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)

Bài 2

\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)

(1-2x)(x-1)-5

=-2x²+3x-1-5

=-2x²+3x-6

=-2(x²-3/2x+3)

=-2(x-3/4)²-39/8

Vì (x-3/4)²≥0  với mọi x

⇒-2(x-3/4)²≤0

⇒-2(x-3/4)²-39/8<0

Vậy biểu thức (1-2x)(x-1)-5 luôn âm với mọi x

Câu trả lời bằng hình ảnh.

\(-3x^2+x-20=-3\left(x^2-\dfrac{1}{3}x+\dfrac{20}{3}\right)\)

\(=-3\left(x^2-2.\dfrac{1}{6}x+\dfrac{1}{36}+\dfrac{239}{36}\right)=-3\left[\left(x-\dfrac{1}{6}\right)^2+\dfrac{239}{36}\right]\)

\(=-3\left(x-\dfrac{1}{6}\right)^2-\dfrac{239}{12}\le-\dfrac{239}{12}< 0\left(\forall x\right)\)

Ta có: \(-3x^2+x-20\)
\(=-3\left(x^2-\dfrac{1}{3}x+\dfrac{20}{3}\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}\right)-\dfrac{239}{12}\)

\(=-3\left(x-\dfrac{1}{6}\right)^2-\dfrac{239}{12}< 0\forall x\)(đpcm)

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 \(\left(x+\frac{1}{5}\right)^2=0\)

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

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

a: \(A=x^2-x+1\)

\(=x^2-x+\frac14+\frac34\)

\(=\left(x-\frac12\right)^2+\frac34\ge\frac34>0\forall x\)

b: \(B=3x^2-2x+5\)

\(=3\left(x^2-\frac23x+\frac53\right)\)

\(=3\left(x^2-2\cdot x\cdot\frac13+\frac19+\frac{14}{9}\right)\)

\(=3\left(x-\frac13\right)^2+\frac{14}{3}>0\forall x\)

c: \(C=x\left(6-x\right)-14\)

\(=6x-x^2-14\)

\(=-x^2+6x-9-5=-\left(x-3\right)^2-5\le-5<0\forall x\)