Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) A = x² - 3x + 5
A = x² - 2.x.\(\dfrac{3}{2}\) + \(\dfrac{9}{4}\) + \(\dfrac{11}{4}\)

A = ( x - \(\dfrac{3}{2}\) )² + \(\dfrac{11}{4}\)

Vì ( x - \(\dfrac{3}{2}\) )² \(\ge\) 0 với mọi x
\(\Rightarrow\) ( x - \(\dfrac{3}{2}\))² + \(\dfrac{11}{4}\) \(\ge\) \(\dfrac{11}{4}\) với mọi x

\(\Rightarrow\) A \(\ge\) \(\dfrac{11}{4}\) với mọi x
Vậy min A = \(\dfrac{11}{4}\) \(\Leftrightarrow\) ( x - \(\dfrac{3}{2}\) )² = 0
\(\Leftrightarrow\) x = \(\dfrac{3}{2}\)

d) D = ( x - 1)( x + 2)( x + 3 )( x + 6)
D = [( x - 1)( x + 6)] [( x + 2 )( x + 3)]
D = ( x² + 6x - x - 6 )( x² + 3x + 2x + 6 )
D = ( x² + 5x - 6 )( x² + 5x + 6)
D = ( x² + 5x )² - 36
Vì ( x² + 5x )² \(\ge\) 0 với mọi x
\(\Rightarrow\) ( x² + 5x )² - 36 \(\ge\) - 36 với mọi x
\(\Rightarrow\) D \(\ge\) -36 với mọi x
Vậy min D = -36 \(\Leftrightarrow\) ( x² + 5x )² = 0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

c) C = x² - 2x + y² - 4y + 7
C = x² - 2x + 1 + y² - 4y + 4 + 2
C = ( x - 1 )² + ( y - 2 )² + 2
Vì ( x - 1 )² \(\ge\) 0 với mọi x
( y - 2 )² \(\ge0\) với mọi y
\(\Rightarrow\)( x - 1)² + ( y - 2)² + 2 \(\ge2\) với mọi x, y
\(\Rightarrow\) C \(\ge2\) với mọi x,y
Vậy min C = 2 \(\Leftrightarrow\) \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

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 \(x=\dfrac{1}{2}\)

h cần hăm

\(a,A=x^2-2x+2=\left(x-1\right)^2+1\ge1\)

dấu"=" xảy ra<=>x=1

\(b,B=2x^2-5x+2=2\left(x^2-\dfrac{5}{2}x+1\right)=2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}-\dfrac{9}{16}\right)\)

\(=2\left[\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{16}\right]=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

dấu"=" xảy ra<=>x=5/4

c,\(C=x^2+2xy+4y^2+3=\left(x+y\right)^2+3\left(y^2+1\right)\ge3\)

dấu"=" xảy ra<=>x=y=0

d,\(D=\left|x-1\right|+|2x-1|=|1-x|+|2x-1|\ge|1-x+2x-1|\)

\(=|x|\ge0\)

dấu"=" xảy ra<=>\(x=0\)

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p