a) \(A=-16x^2-48x-40=-\left(16x^2+48x+36\right)-4\)

\(=-\left(4x+6\right)^2-4\le-4< 0\)

Vậy A vô nghiệm

b) \(B=5x^2+12x+20=5\left(x^2+\dfrac{12}{5}x+\dfrac{36}{25}\right)+\dfrac{64}{5}\)

\(=5\left(x+\dfrac{6}{5}\right)^2+\dfrac{64}{5}\ge\dfrac{64}{5}>0\)

Vậy B vô nghiệm

b: ta có: \(B=5x^2+12x+20\)

\(=5\left(x^2+\dfrac{12}{5}x+4\right)\)

\(=5\left(x^2+2\cdot x\cdot\dfrac{6}{5}+\dfrac{36}{25}+\dfrac{64}{25}\right)\)

\(=5\left(x+\dfrac{6}{5}\right)^2+\dfrac{64}{5}>0\forall x\)

b: Ta có: \(B=5x^2+12x+20\)

\(=5\left(x^2+\dfrac{12}{5}x+4\right)\)

\(=5\left(x+\dfrac{6}{5}\right)^2+\dfrac{64}{5}>0\forall x\)

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 ?

a.

\(P\left(x\right)=x^2-6x+10=x^2-6x+9+1=\left(x-3\right)^2+1>1\forall x\in R\)\(Q\left(x\right)=\left(x-3\right)\left(x-5\right)+4=x^2-8x+15+4=x^2-8x+16+3=\left(x-4\right)^2+3>0\forall x\in R\)b.

\(A\left(x\right)=4x-5-x^2=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left(x-2\right)^2-1< 0\forall x\in R\)\(B\left(x\right)=24x-18-9x^2=-\left(9x^2-24x+18\right)=\left(-9x^2-24x+16+2\right)=-\left(3x+4\right)^2-2< 0\forall x\in R\)

a, P(x) =x^2-6x+10=x^2-6x+9+1=(x+3)^2+1>0

Q(x) =(x-3)(x-5)+4=x^2-8x+15+4=x^2-8x+19=x^2-8x+16+3=(x-4)^2+3>0

Kết luận:với bất kì giá trị nào của biến x thì 2 đa thức trên dương

b, A(x) =4x-5-x^2=-x^2+4x-5=-x^2+4x-4-1=-(x-2)^2-1<0

B(x) =24x-18-9x^2=-9x^2+24x-18= -(3x)^2+24x-16-2=-(3x-4)^2-2<0

Kết luận : ko có giá trị nào của biến x mà 2 đa thức trên dương

a) Ta có: \(m=\left(4x+3\right)^2-2x\left(x+6\right)-5\left(x-2\right)\left(x+2\right)=16x^2+24x+9-2x^2-12x-5\left(x^2-4\right)\)

\(=14x^2+12x+9-5x^2+20=9x^2+12x+29\)

b) \(9x^2+12x+29=\left(9x^2+12x+16\right)+12=\left(3x+4\right)^2+12\ge12\)

Dấu "=" xảy ra khi 3x+4=0 => x=\(\frac{-4}{3}\) => đa thức trên luôn dương.

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

Vì \(\left(x-\frac{1}{2}\right)^2\ge0=>\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\) (với mọi x)

Vậy ........

\(2,a,\left(x-3\right)\left(1-x\right)-2=x-x^2-3+3x-2=-x^2+4x-5=-\left(x^2-4x+5\right)\)

\(=-\left(x^2-4x+4+1\right)=-\left(x^2-2.x.2+2^2+1\right)=-\left[\left(x-2\right)^2+1\right]=-1-\left(x-2\right)^2\)

Vì \(\left(x-2\right)^2\ge0=>-\left(x-2\right)^2\le0=>-1-\left(x-2\right)^2\le-1< 0\) (với mọi x)

Vậy........

\(b,\left(x+4\right)\left(2-x\right)-10=2x-x^2+8-4x-10=-x^2-2x-2=-\left(x^2+2x+2\right)=-\left(x^2+2x+1+1\right)\)

\(=-\left(x^2+2.x.1+1^2+1\right)=-\left(x+1\right)^2+1=-1-\left(x+1\right)^2\le-1< 0\) (với mọi x)

Vậy.......

a) −x2+6x−15=−(x2−6x+15)=−((x−3)2+6)−x2+6x−15=−(x2−6x+15)=−((x−3)2+6)

= −(x−3)2−6−(x−3)2−6 ≤6<0∀x≤6<0∀x (đpcm)

b) (x−3).(1−x)−2=x−x2−3+3x−2=−x2+4x−5(x−3).(1−x)−2=x−x2−3+3x−2=−x2+4x−5

= −(x2−4x+5)−(x2−4x+5) = −((x−2)2+1)=−(x−2)2−1≤−1<0∀x−((x−2)2+1)=−(x−2)2−1≤−1<0∀x (đpcm)

c) (x+4)(2−x)−10=2x−x2+8−4x−10(x+4)(2−x)−10=2x−x2+8−4x−10

−x2−2x−2=−(x2+2x+2)=−((x+1)2+1)=−(x+1)2−1≤−1<0∀x−x2−2x−2=−(x2+2x+2)=−((x+1)2+1)=−(x+1)2−1≤−1<0∀x(đpcm)

a. -x^2+6x-15=-(x^2-6x+9)+9-15=-(x-3)^2-6<=-6<0
b. -9x^2+24x-18=-(9x^2-2.3.4x+16)+16-18=-93x-4)^2-x<=-2<0

a) -x² + 6x - 15

= (-x² + 6x - 9) - 6

= -( x - 3 )² - 6

Vì âm trừ dương ra âm nên ta có :

-x² + 6x -15 luôn âm với mọi x

Phần b mình o biết làlàm

\(a,=\left(5x-1\right)^2\\ b,=\left(x+4\right)^2\\ c,=\left(4x+3y\right)^2\\ d,=\left(\dfrac{x}{4}+2y\right)^2\)