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

Dấu '=' xảy ra khi x=1/2

\(D=25x^2-10xy+y^2+2y^2+4y+2-1\)

\(=\left(5x-y\right)^2+2\left(y+1\right)^2-1>=-1\)

Dấu '=' xảy ra khi y=-1 và x=1/5

Bài 1: 

a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)

\(=x^2-3x+6x-12\)

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

b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

c: \(\left(-2xy+3\right)\left(xy+1\right)\)

\(=-2x^2y^2-2xy+3xy+3\)

\(=-2x^2y^2+xy+3\)

d: \(x\left(xy-1\right)\left(xy+1\right)\)

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

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

Bài 2: 

a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(=27x^3+8\)

\(=27\cdot\dfrac{1}{27}+8=9\)

b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)

\(=125x^3-8y^3\)

\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)

=0

Bài 3: 

Ta có: \(A=\left(x+2\right)\left(3x-1\right)-x\left(3x+3\right)-2x+7\)

\(=3x^2-x+6x-2-3x^2-9x-2x+7\)

=5

\(M=6x^2+4y^2+6xy+\left(xy+\dfrac{4x}{y}\right)+\left(3xy+\dfrac{3y}{x}\right)+2022\)

\(M\ge3x^2+y^2+3\left(x+y\right)^2+2\sqrt{\dfrac{4x^2y}{y}}+2\sqrt{\dfrac{9xy^2}{x}}+2022\)

\(M\ge3\left(x^2+1\right)+\left(y^2+4\right)+3\left(x+y\right)^2+4x+6y+2015\)

\(M\ge6x+4y+3\left(x+y\right)^2+4x+6y+2015\)

\(M\ge3\left(x+y\right)^2+10\left(x+y\right)+2015\ge3.3^2+10.3+2015=2072\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

A = x² - 10x + 12

= ( x² - 10x + 25 ) - 13

= ( x - 5 )² - 13

( x - 5 )² ≥ 0 ∀ x => ( x - 5 )² - 13 ≥ -13

Đẳng thức xảy ra <=> x - 5 = 0 => x = 5

=> MinA = -13 <=> x = 5

B = 6y² + 4y - 1

= 6( y² + 2/3y + 1/9 ) - 5/3

= 6( y + 1/3 )² - 5/3

6( y + 1/3 )² ≥ 0 ∀ x => 6( y + 1/3 )² - 5/3 ≥ -5/3

Đẳng thức xảy ra <=> y + 1/3 = 0 => y = -1/3

=> MinB = -5/3 <=> y = -1/3

C = x² + y² - 2x - 6y - 1

= ( x² - 2x + 1 ) + ( y² - 6y + 9 ) - 11

= ( x - 1 )² + ( y - 3 )² - 11

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y-3\right)^2-11\ge-11\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-1=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

=> MinC = -11 <=> x = 1 ; y = 3

D = 2x² + 3y² - x - 3y + 5

= 2( x² - 1/2x + 1/16 ) + 3( y² - y + 1/4 ) + 33/8

= 2( x - 1/4 )² + 3( y - 1/2 )² + 33/8

\(\hept{\begin{cases}2\left(x-\frac{1}{4}\right)^2\ge0\forall x\\3\left(y-\frac{1}{2}\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x-\frac{1}{4}\right)^2+3\left(y-\frac{1}{2}\right)^2+\frac{33}{8}\ge\frac{33}{8}\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{4}=0\\y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{1}{2}\end{cases}}\)

=> MinD = 33/8 <=> x = 1/4 ; y = 1/2

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

^^

1/

( a + b )³ + ( a - b )³ - 6ab² < đã sửa >

= a³ + 3a²b + 3ab² + b³ + a³ - 3a²b + 3ab² - b³ - 6ab²

= 2a³ 

2/

A = x² + y² - 2x - 4y + 6 = ( x² - 2x + 1 ) + ( y² - 4y + 4 ) + 1 = ( x - 1 )² + ( y - 2 )² + 1 ≥ 1 ∀ x, y

Dấu "=" xảy ra khi x = 1 ; y = 2

=> MinA = 1 <=> x = 1 ; y = 2

B = 2x² + 8x + 10 = 2( x² + 4x + 4 ) + 2 = 2( x + 2 )² + 2 ≥ 2 ∀ x

Dấu "=" xảy ra khi x = -2

=> MinB = 2 <=> x = -2

C = 25x² + 3y² - 10x + 11 = ( 25x² - 10x + 1 ) + 3y² + 10 = ( 5x - 1 )² + 3y² + 10 ≥ 10 ∀ x, y

Dấu "=" xảy ra khi x = 1/5 ; y = 0

=> MinC = 10 <=> x = 1/5 ; y = 0

D = ( x - 3 )² + ( x - 11 )²

Đặt t = x - 7

D = ( t + 4 )² + ( t - 4 )²

    = t² + 8t + 16 + t² - 8t + 16

    = t² + 32 ≥ 32 ∀ t

Dấu "=" xảy ra khi t = 0

=> x - 7 = 0 => x = 7

=> MinD = 32 <=> x = 7

Cảm ơn bn nhiều nhé!

Bài 1:

\(\left(a+b\right)^3+\left(a-b\right)^3-6ab^2\)

\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)-6ab^2\)

\(=2a\left(a^2+3b^2\right)-6ab^2\)

\(=2a^3+6ab^2-6ab^2\)

\(=2a^3\)

Bài 2:

\(A=x^2+y^2-2x-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu"=" xảy ra khi \(\hept{\begin{cases}x-1=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Vậy...

\(B=2x^2+8x+10\)

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

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

Dấu"="xảy ra khi \(x+2=0\Leftrightarrow x=-2\)

Vậy...

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

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

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

d) \(=x^2-10xy+\left(5y\right)^2=\left(x-5y\right)^2\)

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

f) \(=\left(5x\right)^2+20x+4=\left(5x+2\right)^2\)

\(a)x^2-4y^2=(x-2y)(x+2y)\\b)x^2-9y^2=(x-3y)(x+3y)\\c)(2x-1)^2-4y^2=(2x-1-2y)(2x-1+2y)\\d) x^2-10xy+25y^2=(x-5y)^2\\e)9x^2-6x+1=(3x-1)^2\\f)25x^2+20x+4=(5x+2)^2\)