`Answer:`

a. Ta có: \(\frac{a}{b}=\frac{1}{3}\Rightarrow\frac{a}{1}=\frac{b}{3}\)

Đặt \(k=\frac{a}{1}=\frac{b}{3}\Rightarrow\hept{\begin{cases}a=k\\b=3k\end{cases}}\)

\(E=\frac{3a+2b}{4a-3b}\)

\(=\frac{3k+2.3k}{4k-3.3k}\)

\(=\frac{3k+6k}{4k-9k}\)

\(=\frac{9k}{-5k}\)

\(=-\frac{9}{5}\)

b. Thay `a-b=5` vào biểu thức `F`, ta được:

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

\(=\frac{3a-a+b}{2a+b}-\frac{4b+a-b}{a+3b}\)

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

\(=1+1\)

\(=0\)

100

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: ...

4:

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

5: 

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

Bài 1 : 

\(N=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có : \(x+y+z=0\Rightarrow x+y=-z;y+z=-x;x+z=-y\)

hay \(-z.\left(-x\right)\left(-y\right)=-zxy\)

mà \(xyz=2\Rightarrow-xyz=-2\)

hay N nhận giá trị -2 

Bài 2 : 

\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)Đặt \(a=10k;b=3k\)

hay \(\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)

hay biểu thức trên nhận giá trị là 24 

c, Ta có : \(a-b=3\Rightarrow a=3+b\)

hay \(\frac{3+b-8}{b-5}-\frac{4\left(3+b\right)-b}{3\left(3+b\right)+3}=\frac{-5+b}{b-5}-\frac{12+4b-b}{9+3b+3}\)

\(=\frac{-5+b}{b-5}-\frac{12+3b}{6+3b}\)quy đồng lên rút gọn, đơn giản rồi 

1.Ta có:\(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

\(\Rightarrow N=\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(-z\right)\left(-x\right)\left(-y\right)=-2\)

2.Ta có:\(\frac{a}{b}=\frac{10}{3}\Rightarrow\frac{a}{10}=\frac{b}{3}\)

Đặt \(\frac{a}{10}=\frac{b}{3}=k\Rightarrow a=10k;b=3k\)

Ta có:\(A=\frac{3a-2b}{a-3b}=\frac{3.10k-2.3k}{10k-3.3k}=\frac{30k-6k}{10k-9k}=\frac{k\left(30-6\right)}{k\left(10-9\right)}=24\)

Vậy....

\(\dfrac{a}{b}=\dfrac{1}{3}\)

nên b=3a

\(E=\dfrac{3a+2b}{4a-3b}=\dfrac{3a+6a}{4a-9a}=\dfrac{9}{-5}=-\dfrac{9}{5}\)

a-b=5 nên a=b+5

\(F=\dfrac{3\left(b+5\right)-5}{2\left(b+5\right)+b}-\dfrac{4b+5}{b+5+3b}\)

\(=\dfrac{3b+10}{3b+10}-1=1-1=0\)