a) \(G=\frac{\frac{3a}{b}-\frac{2b}{b}}{\frac{a}{b}-\frac{3b}{b}}=\frac{3.\frac{10}{3}-2}{\frac{10}{3}-3}=\frac{10-2}{\frac{1}{3}}=24\)

b) \(H_1=\frac{\frac{2a-3b}{b}}{\frac{4a+3b}{b}}=\frac{\frac{2a}{b}-\frac{3b}{b}}{\frac{4a}{b}+\frac{3b}{b}}=\frac{2.\frac{10}{3}-3}{4.\frac{10}{3}+3}=\frac{\frac{11}{3}}{\frac{49}{3}}=\frac{11}{49}\)

\(H_2=\frac{\frac{5a-4b}{b}}{\frac{3a+b}{b}}=\frac{5.\frac{a}{b}-4}{3.\frac{a}{b}+1}=\frac{5.\frac{10}{3}-4}{3.\frac{10}{3}+1}=\frac{\frac{38}{3}}{\frac{33}{3}}=\frac{38}{33}\)

=> \(H=\frac{11}{49}-\frac{38}{33}=\frac{-1499}{1617}\)

A = 0

A=0

`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:

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

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

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

\(=1+1\)

\(=0\)

a) 3a + 4b - 5c - 2a - 3b + 5c

= ( 3a - 2a ) + ( 4b - 3b ) - ( 5c - 5c )

= a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= ( 7a - 3a - 4a ) + ( 3b + 2b + b ) - ( 4c + 2c + 2c ) 

= 6b - 8c

a) 3a + 4b - 5c - 2a - 3b + 5c

= (3a - 2a) + (4b - 3b) - (5c - 5c)

= a + b - 0 = a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= (7a - 3a - 4a) + (3b + 2b + b) - ( 4c + 2c + 2c)

= 0 + 6b - 8c = 6b - 8c

a)

3a + 4b - 5c - 2a - 3b + 5c

=( 3a - 2a ) + ( 4b - 3b ) + ( -5c + 5c )

= a + b

b)

7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

=( 7a - 3a - 4a ) + ( 3b + 2b + b ) + ( -4c - 2c - 2c )

= 6b + (-8c)

Lời giải:

$\frac{2a-3b}{3a+4b}=\frac{3}{5}$

$\Rightarrow 5(2a-3b)=3(3a+4b)$

$\Rightarrow 10a-15b=9a+12b$

$\Rightarrow 10a-9a=12b+15b$

$\Rightarrow a=27b$

$\Rightarrow \frac{a}{b}=27$

1/Tự chép lại đb nha :v

 =a² - 9b²+2ab+3a²-8b²-12ab+6ab-3b²-2a²+ab

= 2a²-3ab-20b²

= (2a²+5ab) - (8ab+20b²)

= a(2a+5b) - 4b(2a+5b)

=(2a+5b)(a-4b)

câu 2 tương tự nhé :)

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

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

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

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

a-b=6

nên a=b+6

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

\(=\dfrac{3b+18-6}{2b+12+b}-1\)

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

Ta có 2a=3b <=> a=\(\frac{3b}{2}\) 
Lại có 3a+4b=46
Do đó 3x\(\frac{3b}{2}\) +4b=46
<=>\(\frac{9b}{2}\) +\(\frac{8b}{2}\) =46
<=>17b=46x2
<=>b=\(\frac{92}{17}\) 

=>a=3x\(\frac{92}{17}\) :2 
<=>a=\(\frac{138}{17}\)

\(\text{Ta có: }2a=3b\Rightarrow a=\frac{3b}{2}\)
\(\Rightarrow3a+4b=3.\frac{3b}{2}+4b=46\)

\(\Rightarrow\frac{9}{2}b+4b=46\)

\(\Rightarrow b.\left(\frac{9}{2}+4\right)=46\)

\(\Rightarrow b.\frac{17}{2}=46\)

\(\Rightarrow b=46:\frac{17}{2}=\frac{92}{17}\)

Từ đây rồi tính  đc a

Ta có : \(2a=3b\Rightarrow\frac{a}{3}=\frac{b}{2}\Rightarrow\frac{3a}{9}=\frac{4b}{8}\left(1\right)\)   và \(3a+4b=46\left(2\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{3a}{9}=\frac{4b}{8}=\frac{3a+4b}{9+8}=\frac{46}{17}\)

\(\Rightarrow3a=\frac{46.9}{17}=24,3\Leftrightarrow a=24,3:3=8,1\)

\(\Rightarrow4b=\frac{46.8}{17}=21,6\Rightarrow b=21,6:4=5,4\)