\(x^4\cdot y^4=16\Leftrightarrow\left(xy\right)^4=16\Leftrightarrow xy=2\)  (1)

 có:  \(\frac{x}{2}=\frac{y}{9}\Leftrightarrow x=\frac{2y}{9}\)

thay vào (1) đc: 

\(x\cdot y=\frac{2y}{9}\cdot y=\frac{2y^2}{9}=2\)

\(\Rightarrow2y^2=18\Leftrightarrow y^2=9\Leftrightarrow y=3\)và \(y=-3\)

y = 3  <=> x = 2*3/9 = 2/3

y = -3  <=> x = 2*(-3)/9=-2/3

vậy x = 2/3, y = 3

      x = -2/3, y = -3

x=1 ; y=2

x=1;y=2

hoặc

x=-1;y=-2

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x^2+y^2}{3^2+4^2}=\frac{16}{25}\)

chỉ bt tek -_- 

x/3=y/5=x+y/3+5=16/8=2

x/3=2 suy ra x=6

y/5=2 suy ra y=10

x/2=y/3suy ra x/8=y/12

y/4=z/5 suy ra y/12=z/15

x/8=y/12=z/15=x+y-z/8+12-15=10/5=2

x/8=2 suy ra x=16

y/12=2 suy ra y=24

x/15=2 suy ra z=30

a) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=\dfrac{16}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.\dfrac{16}{5}\\y^2=9.\dfrac{16}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm\left(2.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{8\sqrt[]{5}}{5}\\y=\pm\left(3.\dfrac{4}{\sqrt[]{5}}\right)=\pm\dfrac{12\sqrt[]{5}}{5}\end{matrix}\right.\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5}{4}y=\dfrac{5}{4}.\left(\pm\dfrac{12\sqrt[]{5}}{5}\right)=\pm3\sqrt[]{5}\)

b) \(\left|2x+3\right|=x+2\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\3x=-\dfrac{5}{3}\end{matrix}\right.\)

Đính chính

Dòng cuối \(3x=-\dfrac{5}{3}\rightarrow x=-\dfrac{5}{3}\)

\(\frac{x}{2}=\frac{y}{4}=k\)

=>   \(x=2k;\)\(y=4k\)

Theo bài ra ta có:

\(x^4.y^4=16\)

<=>  \(\left(2k\right)^4.\left(4k\right)^4=16\)

<=> \(4096.k^8=16\)

<=> \(k^8=\frac{1}{256}\)

<=>  \(k=\pm\frac{1}{2}\)

làm nốt phần còn lại

        x/2=y/4

=>   2y=4x

<=>   y=2x

thay vào , ta có

        x⁴ .(2x)⁴ =16

<=> 16x⁸=16

<=>    x⁸ =1

=> x= 1 hoặc x=-1

thay vào ta có 2 cặp (x,y) là ( 1,2) và (-1,-2)

\(\Rightarrow\frac{x^8}{256}=\frac{y^8}{65536}=\frac{x^4.y^4}{4096}=\frac{16}{4096}=\frac{1}{256}\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}y=2\\y=-2\end{array}\right.\)

Mà 2 và 4 cùng dấu

=> x; y cùng dấu

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;2\right);\left(-1;-2\right)\right\}\)

=>\(\frac{x}{2}=\frac{y}{4}=>\frac{x^4}{16}=\frac{y^4}{256}=\frac{x^4.y^4}{16.256}=\frac{16}{4096}=\frac{1}{256}\)

=>\(\begin{cases}x=1\\x=-1\end{cases}\)

=>\(\begin{cases}y=2\\y=-2\end{cases}\)

vậy:

\(x=1;y=2\)

\(x=-1;y=-2\)

\(\frac{x}{2}=\frac{y}{4}\) => \(\frac{x^4}{2^4}=\frac{y^4}{4^4}=\left(\frac{x}{2}\right)^4=\left(\frac{y}{4}\right)^4\)

Đặt: \(\left(\frac{x}{2}\right)^4=\left(\frac{y}{4}\right)^4=k\)

=> \(x^4=k.2^4\)

\(y^4=k.4^4\)

\(\left(xy\right)^4=8^4.k^2=4096.k^2=16\) => \(k^2=\frac{1}{256}\)

=> \(k=\frac{\sqrt{1}}{\sqrt{256}}=\frac{1}{16}\)

=> \(x=\sqrt[4]{\frac{1}{16}.2^4}=1\)

\(y=\sqrt[4]{\frac{1}{16}.4^4}=2\)