a,\(\dfrac{9a^2-16b^2}{4b-3a}=\dfrac{\left(3a-4b\right)\left(3a+4b\right)}{\text{4b-3a}}=-3a-4b\)

b,\(\dfrac{25a^2-30ab+9b^2}{3b-5a}=\dfrac{\left(5a-3b\right)^2}{3b-5a}=3b-5a\)

c,\(\dfrac{27a^3-27a^2+9a-1}{9a^2-6a+1}=\dfrac{27a^3-9a^2-18a^2+6a+3a-1}{9a^2-6a+1}=\dfrac{\left(3a-1\right)\left(9a^2-6a+1\right)}{9a^2-6a+1}=3a-1\)

ĐK \(9a^2-b^2\ne0\)

Ta có B =\(\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}=\frac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a+b\right)\left(3a-b\right)}\)

=\(\frac{6a^2+2ab-3ab-b^2+15ab-5b^2-3a^2+ab}{9a^2-b^2}\)

=\(\frac{3a^2+15ab-6b^2}{9a^2-b^2}=\frac{3\left(a^2+5ab-2b^2\right)}{9a^2-b^2}\)

Từ \(10a^2-3b^2+5ab=0\Rightarrow5ab=3b^2-10a^2\)

\(\Rightarrow B=\frac{3\left(a^2+3b^2-10a^2-2b^2\right)}{9a^2-b^2}=\frac{3\left(-9a^2+b^2\right)}{9a^2-b^2}=-3\)

Vậy B =-3

x²(y+z)+y²(z+y)+z²(x+y)

Theo giả thiết, ta có:

\(10a^2-3b^2+5ab=0\)

nên   \(3\left(10a^2-3b^2+5ab\right)=0\)

\(\Leftrightarrow\)  \(30a^2-9b^2+15ab=0\)

\(\Leftrightarrow\)  \(15ab=-30a^2+9b^2\)

Do đó:  \(A=\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}=\frac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}=\frac{3a^2+15ab-6b^2}{9a^2-b^2}=\frac{3a^2+\left(-30a^2+9b^2\right)-6b^2}{9a^2-b^2}\)

             \(A=\frac{-27a^2+3b^2}{9a^2-b^2}=\frac{-3\left(9a^2-b^2\right)}{9a^2-b^2}=-3\)  (do  \(9a^2-b^2\ne0\)  )

1. \(96-3\left(x.1\right)=42\)

\(96-3x=42\)

\(3x=96-42\)

\(3x=54\)

\(x=54:3\)

\(x=18\)

2.Theo bài ra ta có: \(a-b=8\)

a. Ta có: \(7a-7b=7\left(a-b\right)\)

\(=7.8=56\)

Bài 1:

96-3(x.1)=42

3.(x.1)= 96 - 42

3. (x.1)= 54

x.1= 54:3

x.1= 18

x=18:1=18

Vậy x = 18

Bài 2:

a) 7a-7b

Theo đề bài cho a-b=8

⇒ 7.a-7.b = 7. (a-b)=7.8=56

1) Tìm x:

\(96-3\left(x.1\right)=42\Leftrightarrow96-3x=42\)

\(\Leftrightarrow3x=96-42=54\Leftrightarrow x=\dfrac{54}{3}=18\)

2)Ta có: a-b=8

\(a)7a-7b=7\left(a-b\right)=7.8=56\)

\(b)9a-3b+9b-3a=6a+6b=6\left(a+b\right)\)

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

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

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

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

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

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