Ói , hoa mắt chóng mặt nhức đầu ,

2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}=\dfrac{1+2y+1+6y}{18+6x}=\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)

⇒ \(\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{28}\)

⇒\(9+3x=28\)

⇒\(3x=19\)

⇒\(x=\dfrac{19}{3}\)

bạn thay vào là tìm được y

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

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

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

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

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

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

5: \(x^2+2x-4y^2+1\)

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

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

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

6: \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

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

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 \(N=y^2-x^2\)