giúp tui

Thực hiện quy đồng ta có :

9xy−1y=2+3x⇔9−x=2xy+3y9xy−1y=2+3x⇔9−x=2xy+3y

⇔4xy+2x+6y+3=21⇔4xy+2x+6y+3=21

Do x,y nguyên dương nên ta có:

⇔(2x+1)(2x+3)=21⇔\hept{2x+1=32y+3=7⇔\hept{x=1y=2

K mk vs đk ạ

\(\frac{9}{xy}-\frac{1}{y}=2+\frac{3}{x}\Rightarrow9-x=2xy+3y\Rightarrow y=\frac{9-x}{2x+3}\)

\(\Rightarrow2y=\frac{18-2x}{2x+3}=\frac{21}{2x+3}-1\inℕ^∗\Leftrightarrow\frac{21}{2x+3}\inℕ^∗,\frac{21}{2x+3}>1\)

\(\Rightarrow2x+3=1;3;7\Rightarrow x=-1;0;2\)----> Nhận \(x=2\Rightarrow y=\frac{9-x}{2x+3}=1\)

Vậy phương trình có nghiệm nguyên dương: (2;1).

8) \(3a=4b\Rightarrow a=\frac{4}{3}b\) ,  \(6b=5c\Rightarrow c=\frac{6}{5}b\)

\(a+b+c=106\Rightarrow\frac{4}{3}b+b+\frac{6}{5}b=106\Rightarrow b=30\)

\(\Rightarrow a=\frac{4}{3}b=40\) , \(c=\frac{6}{5}b=36\)

9) \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Rightarrow a=b=c\ne0\)

\(Q=\frac{2a+3b+5c}{a+b+c}=\frac{2a+3a+5a}{a+a+a}=\frac{10}{3}\)

10) Lấy nghịch đảo điều kiện của câu 10 rồi ta suy hoàn toàn như câu 9 được \(a=b=c\)

\(G=\left(1+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(5+\frac{c}{a}\right)=\left(1+\frac{a}{a}\right)\left(3+\frac{a}{a}\right)\left(5+\frac{a}{a}\right)=2.4.6=48\)

Ta có : \(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}\)

Đặt \(x=2k;y=5k\Rightarrow xy=40\Rightarrow10k^2=40\Rightarrow k=\pm2\)

Với k = 2 : \(x=4;y=10\)

Với k = -2 : \(x=-4;y=-10\)

 5x=2y tương đương x/2=y/5

Áp dụng...........:

 x/2=y/5=x.y/2.5=40/10=4

x/2=4 suy ra x=8

y/5=4 suy ra x=20

Ta có: 2a = 3b \(\Rightarrow\frac{a}{3}=\frac{b}{2}\) \(\Rightarrow\frac{a}{9}=\frac{b}{6}\)

4b = 3c \(\Rightarrow\frac{b}{3}=\frac{c}{4}\) \(\Rightarrow\frac{b}{6}=\frac{c}{8}\)

\(\Rightarrow\frac{a}{9}=\frac{b}{6}=\frac{c}{8}=\frac{a+b+c}{9+6+8}=\frac{46}{23}=2\)

\(\Rightarrow a=2.9=18;b=2.6=12;c=2.8=16\)

Vậy a = 18; b = 12; c = 2+

thấy ảnh chưa

a) 27² : 25³

= (3³)² : (5²)³

= 3⁶ : 5⁶

 \(=\left(\frac{3}{5}\right)^6\)

b) 25⁴ : 2⁸

= (5²)⁴ : 2⁸

= 5⁸ : 2⁸

\(=\left(\frac{5}{2}\right)^8\)

Ta có:

\(\hept{\begin{cases}\frac{ab}{a+b}=\frac{bc}{b+c}\\\frac{ab}{a+b}=\frac{ca}{c+a}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{a+b}{ab}=\frac{b+c}{bc}\\\frac{a+b}{ab}=\frac{c+a}{ca}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\\\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{a}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=c\\b=c\end{cases}}\)

\(\Rightarrow a=b=c\)