\(a,-x^2+2x+5=-\left(x^2-2x-5\right)=-\left(x^2-2x+1-6\right)=-\left(x-1\right)^2+6\le6\)

dấu'=' xảy ra<=>x=1=>Max A=6

\(b,B=-x^2-y^2+4x+4y+2=-x^2+4x-4-y^2+4x-4+10\)

\(=-\left(x^2-4x+4\right)-\left(y^2-4x+4\right)+10\)

\(=-\left(x-2\right)^2-\left(y-2\right)^2+10=-\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+10\le10\)

dấu"=" xảy ra<=>x=y=2=>Max B=10

\(c,C=x^2+y^2-2x+6y+12=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

dấu'=' xảy ra<=>x=1,y=-3=>MinC=2

a: \(A=x^2+y^2=\left(x+y\right)^2-2xy=15^2-2\cdot50=115\)

c: \(x-y=\sqrt{\left(x+y\right)^2-4xy}=\sqrt{15^2-4\cdot50}=5\)

\(C=x^2-y^2=\left(x+y\right)\left(x-y\right)=15\cdot5=75\)

a) \(\left(x-5\right)^2=\left(3+2x\right)^2\)

\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)

\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)

b) \(27x^3-54x^2+36x=9\)

\(\Rightarrow27x^3-54x^2+36x-9=0\)

\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)

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

\(\Rightarrow\left(3x-2-1\right)\left[\left(3x-2\right)^2+3x-2+1\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)

mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)

\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)

(\(x-5\))² = (3 +2\(x\))² ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}

  27\(x^3\) - 54\(x^2\) + 36\(x\) = 9

27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1

(3\(x\) - 2)³ = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1

1

(x - 5)² = (3 + 2x)²

(x - 5)² - (3 + 2x)² = 0

[(x - 5) - (3 + 2x)][(x - 5) + (3 + 2x)] = 0

(x - 5 - 3 - 2x)(x - 5 + 3 + 2x) = 0

(-x - 8)(3x - 2) = 0

-x - 8 = 0 hoặc 3x - 2 = 0

*) -x - 8 = 0

-x = 8

x = -8

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = -8; x = 2/3

--------------------

27x³ - 54x² + 36x = 9

27x³ - 54x² + 36x - 9 = 0

27x³ - 27x² - 27x² + 27x + 9x - 9 = 0

(27x³ - 27x²) - (27x² - 27x) + (9x - 9) = 0

27x²(x - 1) - 27x(x - 1) + 9(x - 1) = 0

(x - 1)(27x² - 27x + 9) = 0

x - 1 = 0 hoặc 27x² - 27x + 9 = 0

*) x - 1 = 0

x = 1

*) 27x² - 27x + 9 = 0

Ta có:

27x² - 27x + 9

= 27(x² - x + 1/3)

= 27(x² - 2.x.1/2 + 1/4 + 1/12)

= 27[(x - 1/2)² + 1/12] > 0 với mọi x ∈ R

⇒ 27x² - 27x + 9 = 0 (vô lí)

Vậy x = 1

A = x² + y²

= x² - 2xy + y² + 2xy

= (x - y)² + 2xy

= 4² + 2.1

= 16 + 2

= 18

B = x³ - y³

= (x - y)(x² + xy + y²)

= (x - y)(x² - 2xy + y² + xy + 2xy)

= (x - y)[(x - y)² + 3xy]

= 4.(4² + 3.1)

= 4.(16 + 3)

= 4.19

= 76

C = x⁴ + y⁴

= (x²)² + (y²)²

= (x²)² + 2x²y² + (y²)² - 2x²y²

= (x² + y²)² - 2x²y²

= (x² - 2x²y² + y² + 2x²y²)² - 2x²y²

= [(x - y)² + 2x²y²]² - 2x²y²

= (4² + 2.1²)² - 2.1²

= (16 + 2)² - 2

= 18² - 2

= 324 - 2

= 322

a: =>(2x+3)^2-(x-5)^2=0

=>(2x+3+x-5)(2x+3-x+5)=0

=>(x+8)(3x-2)=0

=>x=2/3 hoặc x=-8

b: =>27x^3-54x^2-36x-9=0

=>3x^3-6x^2-4x-1=0

=>\(x\simeq2,57\)

c: A=x^2+y^2=(x-y)^2+2xy=4^2+2=18

B=x^3-y^3=(x-y)^3+3xy(x-y)

=4^3+3*1*4

=64+12=76

C=(x^2+y^2)^2-2x^2y^2

=18^2-2*1^2=322

a) \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)

b) \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x=x^3-3x^2+3x-1-x^3-x^2-x+x^2+x+1-3x+3x^2=0\)

a: Ta có: \(A=y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)

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

=0

b: Ta có: \(B=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)

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

=0

phân tích đa thức thành nhân tử

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

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

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2