a/ \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left[\left(x+1\right)\left(x-6\right)\right].\left[\left(x-2\right)\left(x-3\right)\right]\)

\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

Suy ra Min A = -36 <=> \(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b/ \(B=19-6x-9x^2=-9\left(x-\frac{1}{3}\right)^2+20\le20\)

Suy ra Min B = 20 <=> x = 1/3

a) \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)

\(=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)

\(\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\)

Vì \(\left(x^2-5x\right)^2\ge0\)

=> \(\left(x^2-5x\right)^2-36\ge-36\)

Vậy GTNN của A là -36 khi \(x^2-5x=0\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b) \(B=19-6x-9x^2=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\)

Vì \(-\left(3x+1\right)^2\le0\)

=> \(-\left(3x+1\right)+20\le20\)

Vậy GTLN của B là 20 khi \(x=-\frac{1}{3}\)

Mk làm luôn nhé , không chép đề đâu

B = \(\dfrac{x^2}{x-4}.\dfrac{x^2-8x+4^2}{x}+5\) ( ĐKXĐ : x # 0 ; x # 4)

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

B = \(x\left(x-4\right)+5\)

B = x² - 4x + 5

B = x² - 4x + 4 + 1

B = ( x - 2)² + 1

Do : ( x - 2)² lớn hơn hoặc bằng 0 với mọi x

=> ( x - 2)² + 1 lớn hơn hoặc băng bằng 1 với mọi x

=> B_min = 1 khi và chỉ khi : x = 2

Bạn ơi , giúp tớ câu trên với

a, ĐK: \(\hept{\begin{cases}x+2\ne0\\x\ne0\end{cases}\Rightarrow}\hept{\begin{cases}x\ne-2\\x\ne0\end{cases}}\)

b, \(B=\left(1-\frac{x^2}{x+2}\right).\frac{x^2+4x+4}{x}-\frac{x^2+6x+4}{x}\)

\(=\frac{-x^2+x+2}{x+2}.\frac{\left(x+2\right)^2}{x}-\frac{x^2+6x+4}{x}\)

\(=\frac{\left(-x^2+x+2\right)\left(x+2\right)-\left(x^2+6x+4\right)}{x}\)

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

\(=\frac{-x^3-2x^2-2x}{x}=-x^2-2x-2\)

c, x = -3 thỏa mãn ĐKXĐ của B nên với x = -3 thì 

\(B=-\left(-3\right)^2-2.\left(-3\right)-2=-9+6-2=-5\)

d, \(B=-x^2-2x-2=-\left(x^2+2x+1\right)-1=-\left(x+1\right)^2-1\le-1\forall x\)

Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)

Vậy GTLN của B là - 1 khi x = -1

Thanks bạn ;)

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4

`A=x^2-4x+1/(x^2-4x+4)+5`
`=x^2-4x+4+1/(x^2-4x+4)+1`
Áp dụng BĐT cosi với 2 số dương ta có:
`x^2-4x+4+1/(x^2-4x+4)=(x-2)^2+1/(x-2)^2>=2`
`=>x^2-4x+4+1/(x^2-4x+4)+1>=3`
Dấu "=" xảy ra khi `(x-2)^2=1/(x-2)^2`
`<=>(x-2)^4=1`
`<=>` $\left[ \begin{array}{l}x=3\\x=1\end{array} \right.$
Vậy `min_A=3<=>` $\left[ \begin{array}{l}x=3\\x=1\end{array} \right.$

1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)

\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)

\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)

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

2: \(\left(x^2-y^2\right)\cdot C=-8\)

=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)

=>\(\left(x-y\right)^3=-8\)

=>x-y=-2

=>x=y-2

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

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

\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)

\(=\left(y-1\right)\left(-4y+4\right)+4xy\)

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

\(=-4y^2+8y-4+4y^2-8y\)
=-4

Em cảm ơn ạ.

1)

a) \(5x\left(x^2-3x+\dfrac{1}{5}\right)\)

\(=5x^3-15x^2+x\)

b) \(\left(x-3\right)\left(2x-1\right)\)

\(=2x^2-x-6x+3\)

\(=2x^2-7x+3\)

2)

a) \(3x^2-15xy\)

\(=3x\left(x-5y\right)\)

b) \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

\(=\left(x-3-y\right)\left(x-3+y\right)\)

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

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

\(=x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+1\right)\left(x+2\right)\)

bài 4

vì x²+1 >0 với mọi x , do đó GT của Q luôn xác định với mọi x

Q=\(\dfrac{2x^2-4x+5}{x^2+1}=\dfrac{\left(3x^2+3\right)+\left(2x^2-4x+2\right)}{x^2+1}\)=\(\dfrac{3\left(x^2+1\right)+2\left(x-1\right)^2}{x^2+1}=\dfrac{3\left(x^2+1\right)}{x^2+1}+\dfrac{2\left(x-1\right)^2}{x^2+1}\)=\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\)

Do (x-1)² ≥ 0

=>2(x-1)² ≥ 0

x²+1 ≥ 0

=>\(\dfrac{2\left(x-1\right)^2}{x^2+1}\ge0\)

=>\(3+\dfrac{2\left(x-1\right)^2}{x^2+1}\ge3\)

=> Q ≥ 3

=>GTNN của Q =3 khi

x-1=0

=>x=1

Vậy GTNN của Q =3 khi x=1