3) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.9=90\end{matrix}\right.\)

4) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x}{10}=\dfrac{2y}{6}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\Rightarrow\left\{{}\begin{matrix}x=7.2=14\\y=7.3=21\end{matrix}\right.\)

5) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-z}{5+7-10}=\dfrac{20}{2}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.7=70\\z=10.10=100\end{matrix}\right.\)

6) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{2z}{10}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\Rightarrow\left\{{}\begin{matrix}x=11.3=33\\y=11.4=44\\z=11.5=55\end{matrix}\right.\)

7) \(\Rightarrow\dfrac{x}{12}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{12+6-10}=\dfrac{20}{8}=\dfrac{5}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}.12=30\\y=\dfrac{5}{2}.6=15\\z=\dfrac{5}{2}.10=25\end{matrix}\right.\)

Câu 3:

\(\dfrac{x}{y}=\dfrac{5}{9}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=5\\ \dfrac{y}{9}=10\Rightarrow y=90\)

Câu b:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\dfrac{x}{2}=7\Rightarrow x=14\\ \dfrac{y}{3}=7\Rightarrow y=21\)

Câu c:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-1}{5+7-10}=\dfrac{20}{2}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=50\\ \dfrac{y}{7}=10\Rightarrow y=70\\ \dfrac{z}{10}=10\Rightarrow z=100\)

Câu d:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\dfrac{x}{3}=11\Rightarrow x=3\\ \dfrac{y}{4}=11\Rightarrow y=44\\ \dfrac{z}{5}=11\Rightarrow z=55\)

Câu e:

\(\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10} \)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{8+6-10}=\dfrac{20}{4}=5\)

\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{6}=5\Rightarrow y=30\\ \dfrac{z}{10}=5\Rightarrow z=50\)

\(\dfrac{4}{3}\) = \(\dfrac{20}{4}\)\(x\) - 1

5\(x\) - 1 = \(\dfrac{4}{3}\)

5\(x\)      = \(\dfrac{4}{3}\) + 1

5\(x\)     = \(\dfrac{7}{3}\)

 \(x\)      = \(\dfrac{7}{3}\) : 5

 \(x\)     = \(\dfrac{7}{15}\)

b,

2\(\times\)\(x\) = 3\(\times\) y 

     \(x\) = \(\dfrac{3}{2}\) \(\times\) y

y - \(\dfrac{3}{2}\) \(\times\) y = 5

\(y\) \(\times\) ( 1 - \(\dfrac{3}{2}\)) = 5

\(y\) \(\times\) - \(\dfrac{1}{2}\) = 5

\(y\)            = 5 : (-\(\dfrac{1}{2}\))

\(y\)            = - 10

\(x\)  = y - 5 = -10 - 5 =-15

Ta có \(\frac{x}{3}=\frac{y}{2};\frac{y}{3}=\frac{z}{5}\Leftrightarrow\frac{x}{9}=\frac{y}{6}=\frac{z}{10}\)

\(\Leftrightarrow\frac{x}{9}=\frac{y}{6}=\frac{z}{10}=\frac{x-y+z}{9-6+10}=\frac{20}{13}\)

Từ \(\frac{x}{9}=\frac{20}{13}\Rightarrow x=\frac{20.9}{13}=\frac{180}{13}\)

      \(\frac{y}{6}=\frac{20}{13}\Rightarrow y=\frac{20.6}{13}=\frac{120}{13}\)

      \(\frac{z}{10}=\frac{20}{13}\Rightarrow z=\frac{20.10}{13}=\frac{200}{13}\)

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: \(10^x:5^y=20^y\)

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

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

\(\Leftrightarrow x=2y\)

C

C

Theo tính chất dãy tỉ số bằng nhau:

 \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{-32}{8}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-20\end{matrix}\right.\)

các bạn giải nhanh nhanh giúp tớ nha chiều nay tớ đi học rồi

\(\frac{x}{y}=\frac{20}{3}\Rightarrow x=\frac{20y}{3}\) \(\Rightarrow xy=\frac{20y}{3}.y=\frac{27}{5}\Leftrightarrow y^2=\frac{27}{5}.\frac{3}{20}=\frac{81}{100}\)

\(\Rightarrow y=\frac{9}{10}\)hoặc \(y=-\frac{9}{10}\)

Với \(y=\frac{9}{10}\Rightarrow x=\frac{27}{5}.\frac{10}{9}=\frac{270}{45}=6\)

Với \(y=-\frac{9}{10}\Rightarrow x=\frac{27}{5}.\frac{-10}{9}=-6\)

Vậy \(\left(x;y\right)\in\left\{\left(6;\frac{9}{10}\right);\left(-6;-\frac{9}{10}\right)\right\}.\)

\(\frac{x}{y}=\frac{20}{3}\Rightarrow x=\frac{20y}{3}\) \(\Rightarrow xy=\frac{20y}{3}.y=\frac{27}{5}\Leftrightarrow y^2=\frac{27}{5}.\frac{3}{20}=\frac{81}{100}\)

\(\Rightarrow y=\frac{9}{10}\)hoặc \(y=-\frac{9}{10}\)

Với \(y=\frac{9}{10}\Rightarrow x=\frac{27}{5}.\frac{10}{9}=\frac{270}{45}=6\)

Với \(y=-\frac{9}{10}\Rightarrow x=\frac{27}{5}.\frac{-10}{9}=-6\)

Vậy \(\left(x;y\right)\in\left\{\left(6;\frac{9}{10}\right);\left(-6;-\frac{9}{10}\right)\right\}.\)

\(\text{a)}\)\(2^{x+1}.3^y=2^{2x}.3^x\Leftrightarrow\frac{2^{2x}}{2^{x+1}}=\frac{3^y}{3^x}\)

                                          \(\Leftrightarrow2^{x-1}=3^{y-x}\)

                                           \(\Leftrightarrow x-1=y-x=0\)

                                           \(\Leftrightarrow x=y=1\)

Giải:

Ta có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\)

\(\frac{y}{2}=\frac{z}{5}\Rightarrow\frac{y}{6}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{-20}{25}=\frac{-4}{5}\)

+) \(\frac{x}{4}=\frac{-4}{5}\Rightarrow x=\frac{-16}{5}\)

+) \(\frac{y}{6}=\frac{-4}{5}\Rightarrow y=\frac{-24}{5}\)

+) \(\frac{z}{15}=\frac{-4}{5}\Rightarrow z=-12\)

Vậy bộ số \(\left(x,y,z\right)\) là \(\left(\frac{-16}{5},\frac{-24}{5},-12\right)\)

Ta có:

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\) (1)

\(\frac{y}{2}=\frac{z}{5}\Rightarrow\frac{y}{6}=\frac{z}{15}\) (2)

Từ (1) và (2)

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\) và x+y+z=-20

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=-\frac{20}{25}=-\frac{4}{5}\)

+)\(\frac{x}{4}=-\frac{4}{5}\Rightarrow x=-\frac{4}{5}\cdot4=-\frac{16}{5}\)

+)\(\frac{y}{6}=-\frac{4}{5}\Rightarrow y=-\frac{4}{5}\cdot6=-\frac{24}{5}\)

+)\(\frac{z}{15}=-\frac{4}{5}\Rightarrow z=-\frac{4}{5}\cdot15=-12\)

Vậy \(x=-\frac{16}{5};y=-\frac{24}{5};z=-12\)

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8