Câu 3: 

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

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

=14x+6

b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)

\(=2x^2+14x-3x^2-3x\)

\(=-x^2+11x\)

Câu 2: 

a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)

\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)

\(=-2x^3+3x-4\)

b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)

\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)

\(=2x^2y^2-3x+\dfrac{3}{2}y\)

c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)

\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)

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

d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)

\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)

\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)

e tách ra 10 câu 1 thôi nhé =)

tách ra nhé bn

1: \(\dfrac{4x^3-2x^2-3x+1}{x-2}\)

\(=\dfrac{4x^3-8x^2+6x^2-12x+9x-18+19}{x-2}\)

\(=4x^2+6x+9+\dfrac{19}{x-2}\)

2: \(\dfrac{2x^4-x^3-3x^2-2x}{x-2}\)

\(=\dfrac{2x^4-4x^3+5x^3-10x^2+7x^2-14x+12x-24+24}{x-2}\)

\(=2x^3+5x^2+7x+12+\dfrac{24}{x-2}\)

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

5)\(\dfrac{x^3-2x^2-9x+18}{x-3}=\dfrac{\left(x-2\right)\left(x-3\right)\left(x+3\right)}{x-3}=\left(x+3\right)\left(x-2\right)=x^2+x-6\)

3)\(\dfrac{2x^4-x^3-3x^2-2x}{x-2}=\dfrac{\left(x-2\right)\left(2x^3+3x^2+3x+4\right)-8}{x-2}=\)\(\left(2x^3+3x^2+3x+4\right)\)\(-\dfrac{8}{x-2}\)

b: Xét ΔABE vuông tại A có AH là đường cao ứng với cạnh huyền BE

nên \(BH\cdot BE=AB^2\left(1\right)\)

Xét ΔABC vuông tại B có BH là đường cao ứng với cạnh huyền AC

nên \(AH\cdot AC=AB^2\left(2\right)\)

Từ (1) và (2) suy ra \(BH\cdot BE=AH\cdot AC\)

4*cos(pi/6-a)*sin(pi/3-a)

=4*(cospi/6*cosa+sinpi/6*sina)*(sinpi/3*cosa-sina*cospi/3)

=4*(căn 3/2*cosa+1/2*sina)*(căn 3/2*cosa-1/2*sina)

=4*(3/4*cos^2a-1/4*sin^2a)

=3cos^2a-sin^2a

=3(1-sin^2a)-sin^2a

=3-4sin^2a

=>m=3; n=-4

m^2-n^2=-7

Ta có:

\(\dfrac{1}{cos^2x-sin^2x}+\dfrac{2tanx}{1-tan^2x}=\dfrac{1}{cos2x}+tan2x=\dfrac{1}{cos2x}+\dfrac{sin2x}{cos2x}=\dfrac{1+sin2x}{cos2x}=\dfrac{cos2x}{1-sin2x}\)

\(\Rightarrow P=a+b=2+1=3\)