Theo đề bài : \(a-b=3\Rightarrow a=b+3\).

Thay \(a=b+3\) vào \(A\) ta được : 

\(A=\dfrac{a-8}{b-5}-\dfrac{4a-b}{3a+3}\)

\(=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\)

\(=\dfrac{b-5}{b-5}-\dfrac{4b+12-b}{3b+9+3}\)

\(=1-\dfrac{3b+12}{3b+12}=1-1=0\)

Vậy : Với \(a-b=3\) thì \(A=0.\)

\(a-b=3\\ \Rightarrow a=3+b\)

Thay \(a=3+b\) vào \(A\)

\(A=\dfrac{b+3-8}{b-5}-\dfrac{4.\left(b+3\right)-b}{3.\left(b+3\right)+3}\\ =\dfrac{b-5}{b-5}-\dfrac{4b+12-b}{3b+9+3}\\ =\dfrac{b-5}{b-5}-\dfrac{3b+12}{3b+12}\\ =1-1=0\)

Vậy \(A=0\)

\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

A=\(\dfrac{5}{9}\)

\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{2a-5b}{-14}=\dfrac{a-3b}{-9}=\dfrac{4a+b}{16}=\dfrac{8a-2b}{16}\\ \Leftrightarrow A=\dfrac{-14}{-9}-\dfrac{16}{16}=\dfrac{14}{9}-1=\dfrac{5}{9}\)

A=5/9

\(\dfrac{a}{6}=\dfrac{b}{9}\)

\(\Leftrightarrow9a=6b\)

\(\Rightarrow3a=2b\)(chia cả 2 vế cho 3)

\(\Rightarrow3a-2b=0\Rightarrow\dfrac{3a-2b}{3a+2b}=0\)

Chúc bn học tốt

Ta có: `a/6 = b/9` `-> 9a = 6b`

`-> 3a = 2b`

Vì `3a = 2b` nên `3a - 2b = 0`.

`-> A = (3a - 2b)/(3a + 2b) = 0/(3a + 2b) = 0` 

Vậy giá trị biểu thức `A` là `0`.

b

4:

D=6a+9b=3(2a+3b)=36

5: 

D=15a+5b=5(3a+b)=90

Câu 5:

\(D\left(2\right)=21a+9b-6a-4b\)

\(D\left(2\right)=\left(21a-6a\right)+\left(9b-4b\right)\)

\(D\left(2\right)=15a+5b\)

Mà: \(3a+b=18\Rightarrow b=18-3b\)

\(\Rightarrow D\left(2\right)=15a+5\left(18-3b\right)\)

\(D\left(2\right)=15a+90-15a\)

\(D\left(2\right)=90\)

Vậy: ...

Câu 4:

\(D\left(1\right)=4a+10b-b+2a\)

\(D\left(1\right)=\left(4a+2a\right)+\left(10b-b\right)\)

\(D\left(1\right)=6a+9b\)

Mà: \(2a+3b=12\Rightarrow a=\dfrac{12-3b}{2}\)

\(\Rightarrow D\left(1\right)=6\left(\dfrac{12-3b}{2}\right)+9b\)

\(D\left(1\right)=\dfrac{6\left(12-3b\right)}{2}+9b\)

\(D\left(1\right)=3\left(12-3b\right)+9b\)

\(D\left(1\right)=36-9b+9b\)

\(D\left(1\right)=36\)

Vậy: ...

Câu 3:

Sửa đề: \(C=5a-4b+7a-8b\)

\(C=\left(5a+7a\right)-\left(4b+8b\right)\)

\(C=12a-12b\)

\(C=12\left(a-b\right)\)

\(C=12\cdot8\)

\(C=96\)

Vậy: ...

Biểu thức có giá trị bằng 2 thì: