12,ĐK : x >= 0

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

13, \(\left(\frac{3}{2}x+3y\right)^2=\frac{9}{4}x+\frac{2.3x}{2}.3y+9y^2=\frac{9}{4}x+9xy+9y^2\)

14, ĐK : x ; y >= 0 

 \(\left(\sqrt{2x}+\sqrt{8y}\right)^2=2x+2\sqrt{16xy}+8y=2x+8\sqrt{xy}+8y\)

15, \(\left(x+\frac{1}{6}y+3\right)^2=x^2+\frac{1}{36}y^2+9-\frac{1}{3}xy-y-6x\)

sửa 13 = \(\frac{9}{4}x^2+9xy+9y^2\)

16, \(\left(\frac{1}{2}x-4y\right)^2=\frac{1}{4}x^2-4xy+16y^2\)

17, \(\left(\frac{x}{2}+2y^2\right)\left(\frac{x}{2}-2y^2\right)=\frac{x^2}{4}-4y^4\)

18, \(\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)

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

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

20, \(\left(2x+3\right)^2-\left(x+1\right)^2=\left(2x+3-x-1\right)\left(2x+3+x+1\right)=\left(x+2\right)\left(3x+4\right)\)

Ta có: (x+y)³ = 1³

=> x³ + 3x²y + 3xy² + y³ = 1

=> x³+ y³ + 3xy ( x+y) =1

=> x³ + y³ + 3.(-6).1 =1

=> x³ + y³ =19

Từ x+y=1

=> (x+y)⁵ = 1⁵

=> x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵ = 1

=> x⁵ + y⁵ + 5xy(x³ + y³) +  10x²y²( x + y) = 1

=> x⁵ + y⁵ + 5.(-6) . 19 + 10 . (-6)² . 1 =1

=> x⁵ + y⁵ - 210 = 1

=> x⁵ + y⁵= 1 +210 = 211

Vậy x⁵ + y⁵ = 211

(a+b)(a² - ab + b²) [ a⁶ - (ab)³ + b⁶]

= (a³ + b³) (a⁶ - a³b³ + b⁶)

= (a³ + b³) [(a³)² - a³ b³ + (b³)²]

= a⁹ + b⁹

\(B=2x^2-6x+7\)

\(=2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}+7\)

\(=2\left(x-\frac{3}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

Vậy \(MinB=\frac{5}{2}\Leftrightarrow x=\frac{3}{2}\)

\(C=\left(2x-5\right)^2-4\left(2x-5\right)\)

\(=\left(2x-5\right)\left(2x-5-4\right)=2x-5\)

\(=[\left(2x-5\right)^2-4\left(2x-5\right)+4]-4\)

\(=\left(2x-5-2\right)^2-4\)

\(=\left(2x-7\right)^2-4\ge-4\)

Vậy \(MinC=-4\Leftrightarrow x=\frac{7}{2}\)

(2x-5)^2 -4(2x-5)=(2x-5)^2 -4(2x-5)+4-4=(2x-7)^2 -4>=-4 suy ra C đạt gtnn là -4

2xⁿ⁺¹ (x^n-1 - y^n-1) - y^n-1 (2xⁿ⁺¹ - yⁿ⁺¹)

= 2x²ⁿ -2xⁿ⁺¹ y^n-1 - y^n-1 2xⁿ⁺¹ +y²ⁿ

= 2x²ⁿ + y²ⁿ - 4xⁿ⁺¹ y^n-1

2x²-x+1/3-2/3x=0

6x²-5x+1=0

6x²-3x-2x+1=0

(x-1)(2x-1)=0

x-1=0 và 2x=1

x=1 và x=1/2

\(x\left(2x-1\right)+\frac{1}{3}-\left(\frac{2}{3}\right)x=0\)

\(\Leftrightarrow2x^2-x+\frac{1}{3}-\frac{2x}{3}=0\)

\(\Leftrightarrow2x^2-\frac{5}{3}x+\frac{1}{3}=0\)

\(\Leftrightarrow\frac{6x^2-5x+1}{3}=0\)

\(\Leftrightarrow6x^2-5x+1=0\)

\(\Leftrightarrow6x^2-2x-3x+1=0\)

\(\Leftrightarrow2x\left(3x-1\right)-\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=1\\2x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{1}{2}\end{cases}}\)

Vậy \(S=\left\{\frac{1}{3};\frac{1}{2}\right\}\)

x² - 4x + 2 = ( x² - 4x + 4 ) - 2 = ( x - 2 )² - 2 ≥ -2 ∀ x

Dấu "=" xảy ra <=> x = 2 . Vậy GTNN của bthuc = -2

x^2 - 4x + 2 

= x^2 - 4x + 4 - 2 

= ( x - 2 ) ^2 - 2 

 \(\left(x-2\right)^2\ge0\forall x\)

\(\left(x-2\right)^2-2\ge-2\)   

Dấu = xảy ra khi và chỉ khi 

 x - 2 = 0 

x = 0 + 2 

x = 2 

vậy min = -2 khi và chỉ khi x = 2 

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

\(=x\left(y-z\right)\left(y+z\right)+yz^2-yx^2+zx^2-zy^2\)

\(=\left(y-z\right)\left[x.\left(y+z\right)\right]-x^2\left(y-z\right)-yz\left(y-z\right)\)

\(=\left(y-z\right)\left(xy+xz\right)-x^2\left(y-z\right)-yz\left(y-z\right)\)

\(=\left(y-z\right)\left(xy+xz-x^2-yz\right)\)

\(=\left(y-z\right)\left[\left(xy-x^2\right)+\left(xz-yz\right)\right]\)

\(=\left(y-z\right)\left[x\left(y-x\right)-z\left(y-x\right)\right]\)

\(=\left(y-z\right)\left(y-x\right)\left(x-z\right)\)