a. Ta có: x²+y²-2x+4y+5=0

⇌(x-1)²+(y-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

b. Ta có: 4x²+y²-4x-6y+10=0

⇌ (2x-1)²+(y-3)²=0

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

c.Ta có: 5x²-4xy+y²-4x+4=0

⇌(2x-y)²+(x-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)

d.Ta có: 2x²-4xy+4y²-10x+25=0

⇌ (x-2y)²+(x-5)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)

a) \(x^3-4x=0\)

\(x\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)

b) \(5x\left(3x-2\right)=4-9x^2\)

\(5x\left(3x-2\right)-\left(4-9x^2\right)=0\)

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}}\)

c) \(x^2+7x=8\)

\(x^2+7x-8=0\)

\(x^2+8x-x-8=0\)

\(x\left(x+8\right)-\left(x+8\right)=0\)

\(\left(x+8\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+8=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-8\\x=1\end{cases}}}\)

d) \(2x^2+4y^2+10x+4xy=-25\)

\(x^2+x^2+4y^2+10x+4xy+25=0\)

\(\left(4y^2+4xy+x^2\right)+\left(x^2+10x+25\right)=0\)

\(\left(2y+x\right)^2+\left(x+5\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}2y+x=0\\x+5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{5}{2}\\x=-5\end{cases}}}\)

a) =(2y+x)²

b)=(5-x)²

c) =2x(x²+6x+9)

d) = 3ab(1/4+x+x²)=3ab(x+1/2)²

1) (x-1)² + (x- 4y)² + (y + 2)² +10 -1-4

GTNN = 5

2) tuong tu 

a,  x²+2xy+y²+2x+2y-15

<=> (x+y )²+2(x+y)+1-16

Đặt x+y =a

<=> a²+2a+1-4²

<=> (a+1)²-4²

<=> (a+5)(a-3) =>( x+y+5)(x+y-3)

b, x²-4xy+4y²-2x-4y-35

<=> (x-2y)²-2(x-2y)+1-36

Đặt (x-2y)  =b 

=> b²-2b+1-6²

<=> (b-1)²-6²

<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)

c, 

a,A= x^2+2xy+y^2+2x+2y-15

= (x+y)^2+(x+y)-15

Đặt x+y=a, ta có:

A=a^2+2a-15

  =a^2+2a+1-16

  =(a+1)^2-4^2

  =(a+1+4)(a+1-4)

  =(a+5)(a-3)

Thay a=x+y, ta có: A=(x+y+5)(x+y-3).

b,B= x^2 - 4xy+4y^2-2x-4y-35

   Hình như là sai đề đó bạn. Phải là x^2 - 4xy+4y^2-2x+4y-35 hoặc x^2 - 4xy+4y^2+2x-4y-35 hoặc x^2 + 4xy+4y^2-2x-4y-35 mới đúng đó bạn. Bạn xem lại đi nha.

c,C=6x^4 - 5x^3+8x^2-5x+6

C= x^2(6x^2-5x+8-5/x+6/x^2)

  =x^2(6(x^2+2+1/x^2)-5(x+1/x)-4)

  =x^2(6(x+1/x)^2-5(x+1/x)-4)

Đặt x+1/x=a, ta có:

C=x^2(6a^2-5a-4)

  =x^2(6a^2+3a-8a-4)

  =x^2(2a+1)(3a-4)

Thay a=x+1/x vào là được bạn nhé.

phân tích các đa thức sau thành nhân tử
a) 4x^2 - 4xy + 4y^2

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

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

b) x^2 - 4xy +4y^2

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

\(=\left(x-2y\right)^2\)
c) x^2 + 10x + 25

\(=x^2+2.x.5+5^2\)

\(=\left(x+5\right)^2\)
d)x^2 - 10x + 25

\(=x^2-2.x.5+5^2\)

\(=\left(x-5\right)^2\)
e) 81 - (x+1)^2

\(=9^2-\left(x+1\right)^2\)

\(=\left(9-x-1\right)\left(9+x+1\right)\)
f) 16x^2 - 64 (y + 1)^2

\(=16x^2-8^2\left(y+1\right)^2\)

\(=16x^2-\left(8y+8\right)^2\)

\(=\left(16-8y-8\right)\left(16+8y+8\right)\)

p/s: ko chắc câu cuối đâu :v

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

\(c,x^2+10+25=\left(x+5\right)^2\)

\(d,x^2-10x+25=\left(x-5\right)^2\)

\(e,81-\left(x+1\right)^2=\left(9-x-1\right)\left(9+x+1\right)=\left(8-x\right)\left(10+x\right)\)\(f,16x^2-64\left(y+1\right)^2=\left(4x\right)^2-\left[8\left(y+1\right)\right]^2=\left(4x-8y-8\right)\left(4x+8y+8\right)\)