đặt b=3.a thì E=\(\frac{3a+9a}{4a-12a}=\frac{12a}{-8a}=-\frac{3}{2}\)

`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:

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

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

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

\(=1+1\)

\(=0\)

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!

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