Ta có (x³ + ax² + bx + 4) : (x - 2)² = x + a + 4 dư (4a + b + 12)x - 4(a + 3) 

(x³ + ax² + bx + 4) \(⋮\) (x - 2)² khi \(\hept{\begin{cases}4a+b+12=0\\a+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}b=0\\a=-3\end{cases}}\)

Vậy b = 0 ; a = -3 

\(1,2x^3+3x^2-8x+3\)

\(=2x^3-2x^2+5x^2-5x-3x+3\)

\(=2x^2\left(x-1\right)+5x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(2x^2+5x-3\right)\left(x-1\right)\)

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

\(2,x^3-5x^2+2x+8\)

\(=x^3+x^2-6x^2-6x+8x+8\)

\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)

\(=\left(x^2-6x+8\right)\left(x+1\right)\)

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

\(3,-6x^3+x^2+5x-2\)

\(=-6x^3-6x^2+7x^2+7x-2x-2\)

\(=-6x^2\left(x+1\right)+7x\left(x+1\right)-2\left(x+1\right)\)

\(=\left(-6x^2+7x-2\right)\left(x+1\right)\)

\(=\left(-6x^2-3x-4x-2\right)\left(x+1\right)\)

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

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

\(4,3x^3+19x^2+4x-12\)

\(=3x^3+18x^2+x^2+6x-2x-12\)

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

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

\(=\left(3x-2\right)\left(x+1\right)\left(x+6\right)\)

Vì AB//CD nên ∠A và ∠D là 2 góc trong cùng phía

=> ∠A + ∠D = 180⁰

Mà ∠A - ∠D = 80⁰

=> ∠A = (180⁰ + 80⁰):2= 130⁰

=> ∠D= 130⁰ - 80⁰ = 50⁰

Vi AB//CD => ∠B và ∠C là 2 góc trong cùng phía

=> ∠B + ∠C = 180⁰

Ta có: ∠B + 2∠C - ∠B - ∠C = 250⁰ - 180⁰

=> ∠C = 70⁰

=> ∠B = 180⁰ - 70⁰ = 110⁰

Vậy ∠A= 130⁰; ∠B = 110⁰ ; ∠C = 70⁰; ∠D = 50⁰

Theo định lí tứ giác thì: ∠A+ ∠B+∠C+ ∠D =360⁰

Mà ∠A + ∠B + ∠C= 216⁰ => ∠D = 360⁰-216⁰=144⁰

∠B + ∠C + ∠D = 324⁰ => ∠A = 360⁰ -324⁰= 36⁰

∠C+ ∠D + ∠A = 288⁰ => ∠ B = 360⁰ - 288⁰=72⁰

=> ∠C = 360⁰ - 144⁰- 36⁰ - 72⁰ =108⁰

Vậy ∠A = 36⁰ ; ∠B = 72⁰ ; ∠C = 108⁰ ; ∠D = 144⁰

a, \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=0\)

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

b, \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\Leftrightarrow12x-5=0\Leftrightarrow x=\frac{5}{12}\)

c, \(\left(x-5\right)^2-x\left(x-4\right)=9\Leftrightarrow x^2-10x+25-x^2+4x=9\)

\(\Leftrightarrow-6x+16=0\Leftrightarrow x=\frac{8}{3}\)

d, \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)

\(\Leftrightarrow-5x+21=0\Leftrightarrow x=\frac{21}{5}\)

Ta có 4E = 4x² - 4xy + 12y² - 8x - 40y + 80 

= 4x² - 4xy + y² - 4(2x - y) + 4 + 11y² - 44y + 44 + 32 

= (2x - y)² - 4(2x - y) + 4 + 11(y² - 4y + 4) + 32 

 = (2x - y - 2)² + (y - 2)² + 32 \(\ge32\)

=> 4E \(\ge32\)

=> E \(\ge\)8

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-y-2=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

Vậy Min E = 8 <=> x = y = 2 

x(a+b) - a-b

=x(a+b) - (a+b)

=(x-1)(a+b)

x( a - b ) - a + b = x( a - b ) - ( a - b ) = ( a - b )( x - 1 )

a( x - y ) - x + y = a( x - y ) - ( x - y ) = ( x - y )( a - 1 )

mấy ý kia không sửa được nữa nên nghỉ:)