a:

ĐKXĐ: \(x^2+3x>=0\)

=>x(x+3)>=0

=>\(\left[{}\begin{matrix}x>=0\\x< =-3\end{matrix}\right.\)

 \(\sqrt{16}-\sqrt{x^2+3x}=0\)

=>\(\sqrt{x^2+3x}=\sqrt{16}\)

=>x^2+3x=16

=>x^2+3x-16=0

\(\text{Δ}=3^2-4\cdot1\cdot\left(-16\right)=9+64=73>0\)

Do đó: Phương trình có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{73}}{2}\\x_2=\dfrac{-3+\sqrt{73}}{2}\end{matrix}\right.\)

b:

ĐKXĐ: \(x\in R\)

 \(3x-1-\sqrt{4x^2-12x+9}=0\)

=>\(\sqrt{\left(2x-3\right)^2}=3x-1\)

=>\(\left\{{}\begin{matrix}3x-1>=0\\\left(3x-1\right)^2=\left(2x-3\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(x+2\right)\left(5x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{4}{5}\left(nhận\right)\end{matrix}\right.\)

c:

ĐKXĐ: \(\left\{{}\begin{matrix}x^2-6x+8>=0\\2x^2-10x+11>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\x< =2\end{matrix}\right.\\\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=\dfrac{5+\sqrt{3}}{2}\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=4\end{matrix}\right.\)

 \(\sqrt{2x^2-10x+11}=\sqrt{x^2-6x+8}\)

\(\Leftrightarrow2x^2-10x+11=x^2-6x+8\)

=>\(x^2-4x+3=0\)

=>(x-1)(x-3)=0

=>x=3(loại) hoặc x=1(nhận)

A = y^2 - 4y + 9 = y^2 - 4y + 4 + 5 

= ( y - 2 )^2 + 5 >= 5 

Dấu ''='' xảy ra khi y = 2 

Vậy GTNN A là 5 khi y = 2

B = x^2 - x + 1 = x^2 - x + 1/4 + 3/4 = ( x - 1/2 )^2 + 3/4 >= 3/4

Dấu ''='' xảy ra khi x = 1/2 

Vậy GTNN B là 3/4 khi x = 1/2 

C = 2x^2 - 6x = 2 ( x^2 - 3x + 9 / 4 - 9/4 ) 

= 2 ( x - 3/2 )^2 - 9/2 >= -9/2 

Dấu ''='' xảy ra khi x = 3/2 

Vậy GTNN C là -9/2 khi x = 3/2 

a) Ta có: \(A=y^2-4y+9\)

\(=y^2-4y+4+5\)

\(=\left(y-2\right)^2+5\ge5\forall y\)

Dấu '=' xảy ra khi y=2

b) Ta có: \(B=x^2-x+1\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a: ta có: \(P=x^2+10x+27\)

\(=x^2+10x+25+2\)

\(=\left(x+5\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=-5

Chọn C

đáng lẽ phải là x^2+2x+3 chứ bạn 

y-1=(3x^2+10x+11)/(x^2+2x+3)-1

y-1=(3x^2+10+11-x^2-2x-3)/(x^2+2x+3)

y-1=(2x^2+8x+8)/(x^2+2x+3)

y-1=2(x+2)^2/(x^2+2x+3)>=0

y>=1

=>Min y=1 khi x+2=0 hay x=-2 

y-4=(3x^2+10x+11)/(x^2+2x+3)-4

y-4=(3x^2+10x+11-4x^2-8x-12)/(x^2+2x+3)

y-4=(-x^2+2x-1)/(x^2+2x+3)

y-4=-(x-1)^2/(x^2+2x+3)<=0

y<=4 

=>Max y=4 khi x-1=0 hay x=1 

tau dell biết

mi hỏi ngu như chó vậy, về hỏi cho xem nó có biết ko

đúng thì like giúp mik nha bạn. Thx bạn

\(A=x^2-2x+50\)

\(A=x^2-2x+1+49\)

\(A=\left(x-1\right)^2+49\ge49\)

Dấu "=" xảy ra khi:

\(x=1\)

\(B=12x-x^2\)

\(B=-x^2+12x\)

\(B=-x^2+12x-36+36\)

\(B=-\left(x^2-12x+36\right)+36\)

\(B=-\left(x-6\right)^2+36\le36\)

Dấu "=" xảy ra khi:

\(x=6\)

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

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

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

\(C=\left(x^2-6x+x-6\right)\left(x^2-3x-2x+6\right)\)

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

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

Dấu "=" xảy ra khi:

\(x^2-5x=0\)

\(\Rightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

bn lên mạng nhé!

k mk nhé!

thanks!

#conmiu#