(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.

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

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

Giúp mk với ạ

hơi ngán dạng này :((((

a, \(x^2-3x+5=x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+5=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

b,

\(x^2-\frac{1}{3}x+\frac{5}{4}=x^2-2.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{5}{4}=\left(x-\frac{1}{6}\right)^2+\frac{11}{9}>0\forall x\)

c,

\(x-x^2-3=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}-3=-\left(x-\frac{1}{2}\right)^2-\frac{11}{4}< 0\forall x\)d,

\(x-2x^2-\frac{5}{2}=-2\left(x^2-\frac{1}{2}x+\frac{5}{4}\right)=-2\left(x^2-2.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{5}{4}\right)=-2\left[\left(x-\frac{1}{4}\right)^2+\frac{19}{16}\right]=-2\left(x-\frac{1}{4}\right)^2-\frac{19}{8}< 0\forall x\)P/s : ko chắc lém :)))

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

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

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

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)