Giải

\(\frac{3+x}{5+y}=\frac{3}{5}\)

\(\Leftrightarrow5\left(3+x\right)=3\left(5+y\right)\)

\(\Leftrightarrow15+5x=15+5y\)

\(\Leftrightarrow5x=5y\)( cùng bớt đi 15 )

\(\Leftrightarrow x=y\) ( cùng chia cho 5 )

Mà \(x+y=16\Leftrightarrow x+x=16\Leftrightarrow x=y=8\)

Vậy \(x=y=8\)

Bạn gì đó ơi , hình như bạn làm sai chỗ \(3\left(5+y\right)\)

à rế?sai ở chỗ 3(5+y)kìa,đáng lẽ ra pk bằng 15+3y chứ!

1;

x=-6

2;

x=3

3;

x=-5

4;

3/19

5;

666

1

x=-6

2

x=3

3

x=-5

5

666

2x+5=y(x-3) 2x+5=(x-3)y -(x-3)y+2x+5=0 (x-3)y-2x-5=0​​ x-3=0 => x=3 y-2=0 => y=2

Chúc học tốt!

1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\ \left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\ \left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\ \frac{98}{99}-x=\frac{7}{2}\\ x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)

2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)

1.

a) 2^x+1.3^y=12³

<=>2^x+1.3^y=(2²)³.3³

<=> 2^x+1=2⁶  và 3^y=3³

<=>x+1=6 và y=3

<=>x=5 và y=3

b) 10^x : 5^y=20^y

<=>10^x=20^y.5^y=100^y=(10²)^y

<=>x=2y (Nhiều số lắm chèn)

c) 2^x=4^y-1 

<=>2^x=2^y-2

<=>x=y-2

Mặt khác: 27^y=3^x+8

<=> 3^3y=3^x+8

<=>3y=x+8

<=>3y=(y-2)+8

<=>2y=6

<=>y=3

=>x=y-2=3-2=1

Sửa câu a xíu he

a) 2^x+1 . 3^y=12^x

<=>2^x+1.3^y=2^2x.3^x

<=>2^x+1=2^2x và 3^y=3^x

<=>x=y

và x+1=2x

<=>x=1 (và y=1)

=>Cặp (x;y)=(1;1)

a) Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=\left(2^2\right)^3\cdot3^3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2\cdot3\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)

Vậy: (x,y)=(5;3)

b) Ta có: \(10^x:5^y=20^y\)

\(\Leftrightarrow10^x=20^y\cdot5^y\)

\(\Leftrightarrow10^x=100^y\)

\(\Leftrightarrow x=2y\)

1. Ta có: \(\frac{3+x}{5+y}=\frac{3}{5}\Leftrightarrow\hept{\begin{cases}3+x=3k\\5+y=5k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\left(k-1\right)\\y=5\left(k-1\right)\end{cases}}\)

\(\Rightarrow x+y=3\left(k-1\right)+5\left(k-1\right)=\left(3+5\right)\left(k-1\right)\)

\(\Rightarrow8\left(k-1\right)=16\)

\(\Leftrightarrow k-1=16\div8\)

\(\Leftrightarrow k-1=2\)

\(\Leftrightarrow k=2+1\)

\(\Leftrightarrow k=3\)

\(\Rightarrow\hept{\begin{cases}x=3.3-3=6\\y=5.3-5=10\end{cases}}\)

Vậy x = 6 và y = 10

Với \(\frac{3+x}{5+y}=\frac{3}{5}\Leftrightarrow x=3a;y=5a\left(1\right)\)

Ta có :

\(x+y=3a+5a\)

hay \(16=3a+5a\)

\(\Leftrightarrow16=8a\)

\(\Leftrightarrow a=2\left(2\right)\)

Thay ( 2 ) vào ( 1 ) . Ta có :

\(x=3.2;y=5.2\)

\(\Leftrightarrow x=6;y=10\)

Vậy x = 6; y=10

2. Ta có: \(\frac{x-7}{y-6}=\frac{7}{6}\Leftrightarrow\hept{\begin{cases}x-7=7k\\y-6=6k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\left(k+1\right)\\y=6\left(k+1\right)\end{cases}}\)

\(\Rightarrow x-y=7\left(k+1\right)-6\left(k+1\right)=\left(7-6\right)\left(k+1\right)\)

\(\Rightarrow k+1=-4\)

\(\Leftrightarrow k=-4-1\)

\(\Leftrightarrow k=-5\)

\(\Rightarrow\hept{\begin{cases}x=7.\left(-5\right)+7=-28\\y=6.\left(-5\right)+6=-24\end{cases}}\)

Vậy x = -28, y = -24

\(\dfrac{x+1}{2}=\dfrac{y-5}{3}=\dfrac{z-4}{4}=\dfrac{x+1+y-5-z+4}{2+3-4}\)

\(=\dfrac{7}{1}=7\)

\(\Rightarrow\left\{{}\begin{matrix}x=7.2-1=13\\y=7.3+5=26\\z=7.4+4=32\end{matrix}\right.\)