\(\left(x-\frac{1}{2}\right)^2=x^2-2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2\)

\(=x^2-x+\frac{1}{4}\)

`(x-1/2)^2`

`=x^2 - 2 . x . 1/2 +(1/2)^2`

`= x^2 - x +1/4`

`1,`

`(3xy -x^2 +y) 2/3 x^2y`

`= 3xy . 2/3x^2y -x^2 . 2/3 x^2y + y . 2/3x^2y`

`= 2x^3y^2 - 2/3 x^4y + 2/3 x^2y^2`

`2,`

`(4x^3 - 5xy + 2x)( (-1)/2xy)`

`= 4x^3 . (-1)/2 xy - 5xy . (-1)/2xy + 2x. (-1)/2 xy`

`= -2x^4y + 5/2 x^2y^2 - x^2y`

(2x + 3)(x - 1) + (2x - 3)(1 - x) = 0 

<=> (x - 1)(2x + 3 - 2x + 3) = 0

<=> 6(x - 1) = 0

<=> x - 1 = 0

<=> x = 1

b) (5x - 4)² - 49x² = 0

<=> (5x - 4)² - (7x)² = 0 

<=> (12x - 4)(-2x - 4) = 0

<=> \(\orbr{\begin{cases}12x-4=0\\-2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)

c) x² + 3x - 10 = 0

<=> x² - 2x + 5x - 10 = 0

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

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

<=> \(\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

`1,`

`(2x+3)(x-1) + (2x-3) (1-x)=0`

`<=> (2x+3) (x-1) + (2x-3) (-x+1)=0`

`<=> (2x+3)(x-1) - (2x-3)(x-1)=0`

`<=> (x-1) (2x+3 - 2x+3)=0`

`<=> x-1=0`

`<=> x=1`

Vậy pt có tập nghiệm : `S={1}`

`2,`

`(5x-4)^2 - 49x^2=0`

`<=> (5x-4)^2 - (7x)^2 =0`

`<=> (5x-4 -7x) (5x-4+7x)=0`

`<=> (-2x-4) (12x-4)=0`

`<=> -2x-4=0` hoặc `12x-4=0`

`<=> x=-2` hoặc `x=1/3`

Vậy pt có tập nghiệm : `S={-2; 1/3}`

`3,`

`x^2 +3x-10=0`

`<=> x^2 + 5x-2x-10=0`

`<=> (x^2 + 5x)-(2x+10)=0`

`<=> x (x+5)-2(x+5)=0`

`<=> (x+5)(x-2)=0`

`<=> x+5=0` hoặc `x-2=0`

`<=> x=-5` hoặc `x=2`

Vậy pt có tập nghiệm : `S={-5;2}`

Bài 1:

a, 4x²+6x=2x(2x+3)

b, 12x(x-2y)-9y(x-2y)=3(x-2y)(4x-3y)

c, 3x³-6x²+3x=3x(x²-2x+1)=3x(x-1)²

d, 2x³-2xy²+12x²+18x=2x(x²-y²)+2x(6x+9)=2x(x²+6x+9-y²)

  =2x[(x+3)²-y² ]=2x(x+y+3)(x-y+3)

Bài 2:

a, 5x(x-1)+10x-10=0 <=> 5x(x-1)+10(x-1)=0  <=> 5(x-1)(x+2)=0

   \(\Leftrightarrow\orbr{\begin{cases}5\left(x-1\right)=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)

b,(x+2)(x+3)-2x=6  <=> (x+2)(x+3)-2(x+3)=0  <=> (x+3)(x+2-2)=0  <=> x(x+3)=0 

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

c, \(\left(x-1\right)\left(x-2\right)-2=0\Leftrightarrow x^2-3x+2-2=0\Leftrightarrow x\left(x-3\right)\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

Bài 3

 a, \(x^4y+3x^3y^2+3x^2y^3+xy^4=xy\left(x^3+3x^2y+3xy^2+y^3\right)=xy\left(x+y\right)^3\)

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

hình học 

Bài 1 \(\widehat{D}=360^o-\widehat{A}-\widehat{B}-\widehat{C}=360^o-50^o-120^o-90^o=100^o\)

Bài 2 \(Tc:\widehat{C}+\widehat{D}=360^o-\widehat{A}-\widehat{B}=360^o-50^o-110^o=200^o\)

\(\Rightarrow\widehat{C}=200^o-\widehat{D}\)mà \(\widehat{C}=3\widehat{D}\)nên ta có \(3\widehat{D}=200^o-\widehat{D}\Leftrightarrow4\widehat{D}=200^o\Leftrightarrow\widehat{D}=50^o\Rightarrow\widehat{C}=3.50^o=150^o\)

Bài 4 \(\widehat{C}+\widehat{D}=360^o-90^o-110^o=160^o\)

Áp dụng dãy tỉ số bằng nhau

\(\frac{\widehat{C}}{3}=\frac{\widehat{D}}{5}=\frac{\widehat{C}+\widehat{D}}{3+5}=\frac{160^0}{8}=30^o\)

\(\Rightarrow\frac{\widehat{C}}{3}=30^o\Rightarrow\widehat{C}=30^o.3=90^o\Rightarrow\widehat{D}=160^o-90^o=70^o\)