Ói , hoa mắt chóng mặt nhức đầu ,

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)

=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75

=>x=7; y=5

b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)

=>4x+9y=8 và -8x+3y=5

=>x=-1/4; y=1

c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)

=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5

=>2x-3y=-5,5 và 3x-2y=-4,5

=>x=-1/2; y=3/2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)

=>\(x=\sqrt{2};y=\sqrt{3}\)

1) Ta có: 3x - x² = -(x² - 3x + 9/4) + 9/4 = -(x - 3/2)² + 9/4

Ta luôn có: -(x - 3/2)² \(\le\)0 \(\forall\)x

=> -(x - 3/2)² + 9/4 \(\le\)9/4 \(\forall\)x

Dấu "=" xảy ra <=> x - 3/2 = 0 <=> x = 3/2

Vậy Max của 3x - x² là 9/4 tại x = 3/2

2) Ta có : -(x² + y²) + x + 3y+  10 = -x² - y² + x + 3y + 10 = -(x² - x + 1/4) - (y² -3y + 9/4) + 25/2 = -(x - 1/2)² - (y - 3/2)² + 25/2

Ta luôn có: -(x - 1/2)² \(\le\)0 \(\forall\)x

           -(y - 3/2)² \(\le\)0 \(\forall\)y

=> -(x - 1/2)² - (y - 3/2)² + 25/2 \(\le\)25/2 \(\forall\)x;y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{3}{2}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{2}\end{cases}}\)

Vậy ...

a) x² - 2x - 4y² - 4y

= (x² - 4y²) - (2x + 4y)

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

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

= (x + 2y)[x - 2(y + 1)]

b) x⁴ + 2x³ - 4x - 4

= (x⁴ - 4) + ( 2x³ - 4x)

= (x² - 2)(x² + 2) + 2x(x² - 2)

= (x² - 2)(x² + 2 + 2x)

c) x³ + 2x²y - x -2y

= (x³ - x) + (2x²y - 2y)

= x(x² - 1) + 2y(x² - 1)

= (x + 2y)(x² - 1)

Bài 2;

\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)

\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)

\(x.\left(x-3\right)-x+3=0\)

\(x.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

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

\(\left(x-1\right)^3=0\)( hằng đẳng thức số 5 )

\(\Rightarrow x=1\)

Vậy \(x=1\)