1.

a) \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

b) \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

Bài 1:

a, \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

Vậy \(x=-4\) hoặc \(x=-1\)

b, \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x=3\) hoặc \(x=-2\)

\(x^8+x^4+1\)

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

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

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

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

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

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

Vì dài quá nên mình chỉ có thể trả lời được mấy câu thôi

Bài 1:

27x³ - 8 : (6x + 9x² +4)

= (3x - 2) (9x² + 6x + 4) : (9x² + 6x + 4)

= 3x - 2

Bài 3:

a, 81x⁴ + 4 = (9x²)² + 36x² + 4 - 36x²

= (9x² + 2)² - (6x)²

= (9x² + 6x + 2)(9x² - 6x + 2)

b, x² + 8x + 15 = x² + 3x + 5x + 15

= x(x + 3) + 5(x + 3)

= (x + 3)(x + 5)

c, x² - x - 12 = x² + 3x - 4x - 12

= x(x + 3) - 4(x + 3)

= (x + 3) (x - 4)

Câu 1:

(27x³ - 8) : (6x + 9x² + 4)

= (3x - 2)(9x² + 6x + 4) : (6x + 9x² + 4)

= 3x - 2

Câu 2:

a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)

= 6x² + 33x - 10x - 55 - 6x² - 14x - 9x - 21

= -76

⇒ đccm

b) (2x + 3)(4x² - 6x + 9) - 2(4x³ - 1)

= 8x³ + 27 - 8x³ + 2

= 29

⇒ đccm

Câu 3:

a) 81x⁴ + 4

= (9x²)² + 2²

= (9x² + 2)² - (6x)²

= (9x² - 6x + 2)(9x² + 6x + 2)

b) x² + 8x + 15

= x² + 3x + 5x + 15

= x(x + 3) + 5(x + 3)

= (x + 3)(x + 5)

c) x² - x - 12

= x² - 4x + 3x - 12

= x(x - 4) + 3(x - 4)

= (x - 4)(x + 3)

Bài 1:

a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)

b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???

d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)

Bài 2:

a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)

b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)

c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)

1. a,\(A=x^2-2x+5=x^2-2.x.1+1^2-1+5\)

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

Do \(\left(x-1\right)^2\ge0\) với \(\forall x\) \((\)dấu "=" xảy ra \(\Leftrightarrow x=1)\)

\(\Rightarrow\left(x-1\right)^2+4\ge4\) hay \(A\ge4\) \((\) dấu "=" xảy ra \(\Leftrightarrow x=1)\)

Vậy Min A=4 tại x=1

b,\(B=2x^2-6x=2\left(x^2-3x\right)\)

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

\(=2.\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\right]\)

\(=2.\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

Do \(2.\left(x-\dfrac{3}{2}\right)^2\ge0\) với mọi x (dấu "=" xảy ra <=> x=\(\dfrac{3}{2}\))

\(\Rightarrow2.\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\) hay \(B\ge-\dfrac{9}{2}\)

(dấu "=" xảy ra <=> x=\(\dfrac{3}{2}\))

Vậy Min B = \(-\dfrac{9}{2}\) tại x=\(\dfrac{3}{2}\)

Bài 2

a,\(A=6x-x^2+3=-\left(x^2-6x-3\right)\)

\(=-\left(x^2-2.x.3+3^2-9-3\right)\)

\(=-\left[\left(x-3\right)^2-12\right]\)

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

Do \(-\left(x-3\right)^2\le0\) với mọi x (dấu "=" xảy ra <=> x=3)

\(\Rightarrow-\left(x-3\right)^2+12\le12\) hay \(A\le12\) (dấu "=" xảy ra <=> x=3)

Vậy Max A =12 tại x=3

b,\(B=x-x^2+2=-\left(x^2-x-2\right)\)

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

\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\)

Do \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\) hay \(B\le\dfrac{9}{4}\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))

Vậy Max B=\(\dfrac{9}{4}\) tại x=\(\dfrac{1}{2}\)

c,\(C=5x-x^2-5=-\left(x^2-5x+5\right)\)

\(=-\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{25}{4}+5\right]\)

\(=-\left[\left(x-\dfrac{5}{2}\right)^2-\dfrac{5}{4}\right]\)

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

Do \(-\left(x-\dfrac{5}{2}\right)^2\le0\) với mọi x (dấu "=" xảy ra <=> x=\(\dfrac{5}{2}\))

\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\) hay \(C\le\dfrac{5}{4}\) (dấu ''='' xảy ra <=> x=\(\dfrac{5}{2}\))

Vậy Max C=\(\dfrac{5}{4}\) tại x=\(\dfrac{5}{2}\)

Mình làm tiếp phần của Dũng Nguyễn nha.

b) \(4x-x^2-5\)

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

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

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

Vì \(-\left(x-2\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-2\right)^2-1\le-1\)

\(\Rightarrow-\left(x-2\right)^2-1< 0\) với mọi x

Vậy \(4x-x^2-5< 0\) với mọi x

c) \(x^2-x+1\)

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

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

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) với mọi x

Vậy \(x^2-x+1>0\) với mọi x

d) \(-x^2+2x-4\)

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

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

\(=-\left(x-1\right)^2-3\)

Vì \(-\left(x-1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-1\right)^2-3\le-3\)

\(\Rightarrow-\left(x-1\right)^2-3< 0\)

Vậy \(-x^2+2x-4< 0\) với mọi x

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

\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

=> đpcm

b. \(4x^2+5x+7\)

\(=\left(2x\right)^2-2.2x.\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{87}{16}\)

= \(\left(2x+\dfrac{5}{4}\right)^2\) + \(\dfrac{87}{16}\) \(\ge\dfrac{87}{16}\)

=> đpcm

Bài 1:

Ta có: \(4x-x^2-5\)

\(=-x^2+4x-5=-x^2+4x-4-1\)

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

Vì \(-\left(x-2\right)^2< 0\forall x\)

\(\Rightarrow-\left(x-2\right)^2-1< 0\forall x\)

\(\Rightarrow4x-x^2-5< 0\forall x\)

Bài 1:

\(4x-x^2-5\)

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

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

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

Vì \(-\left(x-2\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-2\right)^2-1\le-1\)

\(\Rightarrow4x-x^2-5< 0\) với mọi x

Bài 2:

a) \(M=x^2+y^2-x+6y+10\)

\(M=x^2-2.x\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+y^2+2.y.3+9-9+10\)

\(M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\left(y+3\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\) với mọi x và y

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow Mmin=\dfrac{3}{4}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

b) \(Q=2x^2-6x\)

\(Q=2\left(x^2-3x\right)\)

\(Q=2\left(x^2-2.x\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}\right)\)

\(Q=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\)

Vì \(2\left(x-\dfrac{3}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(\Rightarrow Qmin=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)