\(A=\frac{9a^5-ab^4-18a^4b+2b^5}{3a^2b^2+ab^4-6a^2b^3-2b^5}\)

\(=\frac{a\left(9a^4-b^4\right)-2b\left(9a^4-b^4\right)}{ab^2\left(3a^2+b^2\right)-2b^3\left(3a^2+b^2\right)}\)

\(=\frac{\left(9a^4-b^4\right)\left(a-2b\right)}{\left(3a^2+b^2\right)\left(ab^2-2b^3\right)}\)

\(=\frac{\left(3a^2-b^2\right)\left(3a^2+b^2\right)\left(a-2b\right)}{\left(3a^2+b^2\right)b^2\left(a-2b\right)}\)

\(=\frac{3a^2-b^2}{b^2}\)

\(=3.\left(\frac{a}{b}\right)^2-1=3.\left(\frac{2}{3}\right)^2-1=\frac{1}{3}\)

a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)

\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)

\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)

\(=2a^2-b^2\)

b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)

\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)

\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)

\(=-7ab+b^2\)

c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)

\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)

\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)

\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)

\(=-7bx+3b^2+2x^2\)

d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)

\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)

\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)

\(=-5ax+32x^2-30a^2\)

a: =2ab+8a^2-b^2-4ab+2ab-6a^2

=2a^2-b^2

b: =6a^2-9ab-4ab+6b^2-6a^2+6ab

=-7ab+6b^2

c: =10bx-5b^2-16bx+8b^2+2x^2-xb

=3b^2+2x^2-7xb

d: =2xa+30x^2+5ax+2x^2-30a^2-12ax

=32x^2-30a^2-5ax

A)Ta có: (3a + 4b) ⋮ 7 ⇒ 2 . (3a + 4b) ⋮ 7 ⇒ (6a + 8b) ⋮ 7 (1)

Ta lại có:

(6a + 8b) + (a + 6b)

=(6a + a) + (8b + 6b)

=7a + 14b

=7a + 7 . 2 . b

=7 . (a + 2b) ⋮ 7 (vì 7 ⋮ 7)

⇒(6a + 8b) + (a + 6b) ⋮ 7 mà (6a + 8b) ⋮ 7 (theo (1))

⇒(a + 6b) ⋮ 7 (ĐPCM)

Vậy...

Xin lỗi anh nhưng câu B) em không hiểu lắm ạ!

Ta có: a-b=6 => a=6+b thế vào BT trên ta có:

D=\(\frac{3\left(6+b\right)-6}{2\left(6+b\right)+b}-\frac{4b+6}{6+b+3b}\)

= \(\frac{18+3b-6}{12+2b+b}-\frac{4b+6}{6+4b}\)

= \(\frac{3b+12}{3b+12}-\frac{4b+6}{4b+6}\)

= 1-1 =0

Cảm ơn bạn nhé!

Ta có \(\hept{\begin{cases}3a=4b\\2b=5c\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{b}{3}=\frac{a}{4}\\\frac{b}{5}=\frac{c}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{b}{15}=\frac{a}{20}\\\frac{b}{15}=\frac{c}{6}\end{cases}}\Leftrightarrow\frac{a}{20}=\frac{b}{15}=\frac{c}{6}\)

Đặt \(\frac{a}{20}=\frac{b}{15}=\frac{c}{6}=k\Leftrightarrow\hept{\begin{cases}a=20k\\b=15k\\c=6k\end{cases}}\)

Khi đó a² + b² + c² = 661

<=> (20k)² + (15k)² + (6k)² = 661

<=> 661k² = 661

<=> k² = 1

<=> k = \(\pm1\)

Khi k = 1 => a = 20 ; b = 15 ; c = 6

Khi k = -1 => a = -20 ; b = - 15 ; c = -6

Ta có \(2a=3b=4c\Leftrightarrow\frac{2a}{12}=\frac{3b}{12}=\frac{4c}{12}\Leftrightarrow\frac{a}{6}=\frac{b}{4}=\frac{c}{3}\)

Áp dụng dãy tỉ số bằng nhau ta có : 

\(\frac{a}{6}=\frac{b}{4}=\frac{c}{3}=\frac{3a}{18}=\frac{4b}{16}=\frac{3a+4b-c}{18+16-3}=\frac{72}{31}\)

=> \(\hept{\begin{cases}a=\frac{432}{31}\\b=\frac{288}{31}\\c=\frac{216}{31}\end{cases}}\)

Ta có : 3a = 2b ==> b = 2/3 x a

Thay b = 2/3 x a vào M ta có :

M=2a - 3b / 3a+2b = 2a - 3. 2/3 .a / 3a + 2. 2/3 .a = 2a - 2a / 3a + 2 . 2/3 . a = 0 / 3a + 2 . 2/3 . a = 0

a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)

\(\Rightarrow\left(b+d\right)c=\left(a+c\right)d\)

\(\Rightarrow dpcm\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2a}{2b}=\dfrac{c}{d}=\dfrac{2a+c}{2b+d}=\dfrac{2a-c}{2b-d}\)

\(\Rightarrow\left(2b-d\right)\left(2a+c\right)=\left(2a-c\right)\left(2b+d\right)\)

\(\Rightarrow dpcm\)

c) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{3a}{3b}=\dfrac{5c}{5d}=\dfrac{3a+5c}{3b+5d}=\dfrac{a-3c}{b-3d}\)

\(\Rightarrow\left(b-3d\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)

\(\Rightarrow dpcm\)

Đính chính câu c

\(\Rightarrow\left(3a+5c\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)