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

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

d) \(3x^2-7x+4\)

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

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

\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)

\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)

f) \(x^2+2xy+y^2-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

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

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

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

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

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

d) \(3x^2-7x+4\)

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

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

\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)

\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)

f) \(x^2+2xy+y^2-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

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

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

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

Phương trình đã cho tương đương 

\(\Leftrightarrow\left|\left(x-y\right)^2+2\left(x-y\right)+1+x-2\right|+\left|x^2-3x+2\right|=2\left(x-2\right)\)  (1) 

Vế trái không âm => x \(\ge\)2 

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

\(\Leftrightarrow\left(x-y+1\right)^2+\left(x-2\right)+\left(x-2\right)\left(x-1\right)=2\left(x-2\right)\)  \(\left(x\ge2\right)\)

\(\Leftrightarrow\left(x-y+1\right)^2+\left(x-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-y+1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)Thỏa mãn điều kiện \(\ge\)2 

Vậy pt có nghiệm \(\hept{\begin{cases}x=2\\y=3\end{cases}}\)

mày thử thay x=2 với y=3 vào pt đi xem đúng ko :)) thằng óc lz ngu vclll :))

có  m`  óc loz í . t cộng đúng 100% rr như m` ấn máy tính còn sai còn đi bảo ng` khác oc loz

giúp mk nha rồi mk tích cho

Bài 2:
Đặt x²-x=0

=>x(x-1)=0

=>x=0 hoặc x=1

a) \(2x^2-2y^2\)

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

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

b) \(x^2-4x+4\)

\(=x^2-2\cdot x\cdot2+2^2\)

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

c) \(x^2+2x+1-y^2\)

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

\(=\left(x-y+1\right)\left(x+y+1\right)\)

d) \(x^2-4x\)

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

e) \(x^2+10x+25\)

\(=x^2+2\cdot x\cdot5+5^2\)

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

g) \(x^2-2xy+y^2-9\)

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

\(=\left(x-y-3\right)\left(x-y+3\right)\)

h) \(2x^2-2\)

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

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

i) \(5x^2-5xy+9x-9y\)

\(=5x\left(x-y\right)+9\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+9\right)\)

k) \(y^2-4y+4-x^2\)

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

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

l) \(x^2-16\)

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

\(=\left(x-4\right)\left(x+4\right)\)

m) \(3x^2-3xy+2x-2y\)

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

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

o) \(3x^4-6x^3+3x^2\)

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

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

a) 2x² - 2y²

 = (2x - 2y)(2x + 2y)

 = 4(x - y)(x + y)

b) x² - 4x + 4

 = (x - 2)²

c) x² + 2x + 1 - y²

 = (x + 1)² - y²

 = (x + 1 - y)(x + 1 + y)

d) x² - 4x 

 = x(x - 4)

e) x² +10x + 25

 = (x + 5)²

g) x² - 2xy + y² - 9

= (x - y)² - 3²

 = (x - y - 3)(x - y + 3)

h) 2x² - 2

= 2(x² - 1) 

 = 2(x - 1)(x + 1)

i) 5x² - 5xy + 9x - 9y

  = 5x(x - y) + 9(x- y)

 = (5x + 9)(x - y)

k) y² - 4y + 4 - x²

 = (y - 2)² - x²

 = (y - 2 - x)(y - 2 + x)

l) x² - 16

 = x² - 4²

 = (x - 4)(x + 4)

m) 3x² - 3xy + 2x -2y

 = 3x(x - y) +2(x-y)

 = (3x + 2)(x - y)

o) 3x⁴ - 6x³ + 3x²

 = 3x⁴ - 3x³ - 3x³ + 3x²

 = 3x³(x - 1) - 3x²(x - 1)

 = (3x³ - 3x²)(x - 1)

 = 3x²(x - 1)(x - 1)

 = 3x².(x - 1)²

\(|x^2+1|-(x^2-4x+4)=3x\\\Rightarrow x^2+1-x^2+4x-4=3x(\text{vì }x^2 + 1 > 0 \forall x )\\\Leftrightarrow 4x-3=3x\\\Leftrightarrow4x-3x=3\\\Leftrightarrow x=3\)

Vậy nghiệm của phương trình là \(x=3\).

Do \(x^2+1>0;\forall x\Rightarrow\left|x^2+1\right|=x^2+1\)

Phương trình trở thành:

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

\(\Leftrightarrow4x-3=3x\)

\(\Leftrightarrow x=3\)

MINH CUNG VAY

Từ \(x^2-2xy+2y^2-2x+6y+5=0\)

\(\Rightarrow\left(x^2-2xy-2x+y^2+2y+1\right)+\left(y^2+4y+4\right)=0\)

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

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

Thay vào P ta có: \(P=\frac{3x^2y-1}{4xy}=\frac{3\cdot\left(-1\right)^2\cdot\left(-2\right)-1}{4\cdot\left(-1\right)\cdot\left(-2\right)}=-\frac{7}{8}\)

cha biet