C=[(x+1)(x-6)][(x-2)(x-3)]

=(x²-5x-6)(x²-5x+6)

=(x²-5x)²-36>=-36

GTNN cua C=-36 tai x²-5x=0=>x(x-5)=0=>x=0 hoac x=5

B=(x-3)²+(x-11)²

  =x²-6x+9+x²-22x+121

  =2x²-28x+130

  =2(x²-14x+65)

  =2(x²-2.7x+7²-7²+65)

  =2[(x-7)²-49+65]

  =2(x-7)²+32

=> vì 2(x-7)² >= 0 

=>2(x-7)²+32 >= 32

=> GTNN của B=32. Khi x=7

A=25x²+3y²-10x+11

  =25x²-10x+1+3y²+10

  =(5x-1)²+3y²+10

 =>vì (5x-1)²+3y² >= 0

 =>(5x-1)²+3y²+10>= 10

 => GTNN của A=10 . Khi x=1/5 và y=0

\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)

b:

\(D=-25x^2+10x-1-10\)

\(=-\left(25x^2-10x+1\right)-10\)

\(=-\left(5x-1\right)^2-10< =-10\)

Dấu = xảy ra khi x=1/5

\(E=-9x^2-6x-1+20\)

\(=-\left(9x^2+6x+1\right)+20\)

\(=-\left(3x+1\right)^2+20< =20\)

Dấu = xảy ra khi x=-1/3

\(F=-x^2+2x-1+1\)

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

Dấu = xảy ra khi x=1

a. \(A=\left[\frac{1}{3}+\frac{3}{x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(9-x^2\right)}+\frac{1}{x+3}\right]\)

\(=\left[\frac{x.\left(x-3\right)}{3.x.\left(x-3\right)}+\frac{3.3}{x\left(x-3\right).3}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{1}{x+3}\right]\)

\(=\left[\frac{x^2-3x+9}{3x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{\left(3-x\right).3}{\left(x+3\right).\left(3-x\right).3}\right]\)

\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}:\left[\frac{x^2+9-3x}{3.\left(3-x\right)\left(3+x\right)}\right]\)

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

\(=\frac{-\left(x-3\right)\left(3+x\right)}{x-3}=-\left(3+x\right)\)

b. Để A < -1 thì:

-(3+x) < -1

=> -3 - x < -1

=> x < -3 - (-1) = -2

Vậy x < -2 thì A < -1.

\(Q=\frac{x^2+2x+1}{x+2}=\frac{\left(x+1\right)^2}{x+2}\ge0\forall x>-2\) có GTNN là 0

=> x . 3,6=27 . -2

=>x . 3,6=-54

=>x=-15

=>x.36=27.-2

=> x.3.6=-54

=> x=-15

x^=27

a)\(-2\left(x+6\right)+6\left(x-10\right)=8.\)

\(\Leftrightarrow-2x-12+6x-60=8\)

\(\Leftrightarrow4x=80\)

\(\Rightarrow x=20\)

b/ \(-4\left(2x+9\right)-\left(-8x+3\right)-\left(x+13\right)=0\)

\(\Leftrightarrow-8x-36+8x-3-x-13=0\)

\(\Leftrightarrow-x=52\)

\(\Rightarrow x=-52\)

c/\(7x\left(2+x\right)-7x\left(x+3\right)=14\)

\(\Leftrightarrow14x+7x^2-7x^2-21x=14\)

\(\Leftrightarrow-7x=14\)

\(\Rightarrow x=-2\)

minh cma on nha

tick đi giải cho

đáp án là a=4.............