\(\frac{x}{4}=\frac{y}{8}=k\)

=>   \(x=4k;\)\(y=8k\)

Ta có:  \(x.y=128\)

<=>  \(4k.8k=128\)

<=>  \(32.k^2=128\)

<=>  \(k^2=4\)

<=> \(k=\pm2\)

đến đây bn thay vào và tính nha

=> \(8x=4y\)

mà x . y = 128 => y = 128 : x 

=> 512 : x = 8x 

=> 512 : 8 = x . x

=> 64 = x^2

=> x = 8 hoặc x = -8

Th1 : x = 8                Th2 : x = -8

=> y = 128 : 8                => y = 128 : ( -8 )

y = 16                                  y = -16

Vậy x = 8 thì y = 16

       x = -8 thì y = -16

Ta có\(\frac{x}{2}=\frac{y+4}{8}\)=> 8x=2(y+4) => 4x=y+4 => y=4x-4=4(x-1) (1)

Lại có xy=8 (2)

Thay (1) vào (2) ta được: x.4(x-1)=8 =>x(x-1)=2 => x² - x =2 => x² -x -2 =0 => x² -2x + x -2=0 => x(x-2) +(x-2)=0

                                                                                                                                          => (x+1)(x-2)=0

                                                                                                                                          => x+1=0 hoặc x-2=0

                                                                                                                                          => x= -1 hoặc x=2

Từ đó suy ra y=4(x-1)=4[(-1) -1]= -8 hoặc y=4(x-1)=4(2-1)=4

Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)

           \(\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\left(2\right)\)

Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)

Ta có : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=\frac{3x}{27}=\frac{2y}{24}=\frac{5z}{50}=\frac{3x-2y+5z}{27-24+50}=\frac{86}{53}\) (đề sai)

b) Đặt : k = \(\frac{x}{5}=\frac{y}{7}\)

=> k² \(=\frac{x}{5}.\frac{y}{7}=\frac{xy}{35}=\frac{140}{35}=4\)

=> k = -2;2

+ k = 2 thì \(\frac{x}{5}=2\Rightarrow x=10\)

                 \(\frac{z}{7}=2\Rightarrow z=14\)

+ k = -2 thì \(\frac{x}{5}=2\Rightarrow x=-10\)

                 \(\frac{z}{7}=2\Rightarrow z=-14\)

Vậy................................

d) \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)

=> \(\frac{y+z-x}{4+6-2}=\frac{8}{8}=1\)

=> \(\frac{x}{2}=1\Rightarrow x=2\)

=> \(\frac{y}{4}=1\Rightarrow y=4\)

=> \(\frac{z}{6}=1\Rightarrow z=6\)

b) \(\frac{x}{3}=\frac{y}{4}\Rightarrow x=y.\frac{3}{4}\)

\(\frac{y}{6}=\frac{z}{8}\Rightarrow z=y.\frac{8}{6}=y.\frac{4}{3}\)

=> \(3x-2y-z=y.3.\frac{3}{4}-2y-y.\frac{4}{3}=13\)

=> \(y.\frac{9}{4}-2y-y.\frac{4}{3}=y.\left(\frac{9}{4}-2-\frac{4}{3}\right)=13\)

=> \(y.\frac{-13}{12}=13\)

\(y=13:\frac{-13}{12}\)

\(y=-12\)

=> \(x=y.\frac{3}{4}=-9\)

=> \(z=y.\frac{4}{3}=-16\)

a) Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)=> x=3k ; y=4k ; z=5k

Ta có:

2x + 3y + 5z = 86

=> 2(3k) + 3(4k) + 5(5k) = 86

     6k + 12k + 25k = 86

     (6 + 12 + 25)k = 86

     43k = 86

     k = 86 : 43 = 2

Vậy x = 3k = 3 . 2 = 6

       y = 4k = 4 . 2 = 8

       z = 5k = 5 . 2 = 10

b) Ta có:

\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\)

\(\frac{y}{6}=\frac{z}{8}\Rightarrow\frac{y}{12}=\frac{z}{16}\)

Vậy \(\frac{x}{9}=\frac{y}{12}=\frac{z}{16}\)

Đặt \(\frac{x}{9}=\frac{y}{12}=\frac{z}{16}=k\)=> x=9k ; y=12k ; z=16k

Ta có:

3x - 2y - z = 13

=> 3(9k) - 2(12k) - 16k = 13

     27k - 24k - 16k = 13

     (27 - 24 - 16)k = 13

      (-13)k = 13

       k = 13 : (-13) = -1

Vậy x = 9k = 9 . (-1) = -9

       y = 12k = 12 . (-1) = -12

       z = 16k = 16 . (-1) = -16

c) Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)=> x=2k ; y=3k ; z=4k

Ta có: xy + yz + zx = 104

=> (2k)(3k) + (3k)(4k) + (4k)(2k) = 104

     6k² + 12k² + 8k² = 104

     (6 + 12 + 8)k² = 104

     26k² = 104

     k² = 104 : 26 = 4

=> k\(\in\){-2;2}

Vậy:

TH1:                                  TH2:

x = 2k = 2 . (-2) = -4            x = 2k = 2 . 2 = 4

y = 3k = 3 . (-2) = -6            y = 3k = 3 . 2 = 6

z = 4k = 4 . (-2) = -8            z = 4k = 4 . 2 = 8

d) Ta có: \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\) và y+z-x=8

Theo tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{y+z-x}{4+6-2}=\frac{8}{8}=1\)

Vì \(\frac{x}{2}\)=1 => x=2.1=2

    \(\frac{y}{4}\)=1 => y=4.1=4

    \(\frac{z}{6}\)=1 => z=6.1=6

a/ \(\left\{{}\begin{matrix}x^2+y+xy\left(x^2+y\right)+xy+1=-\frac{1}{4}\\x^4+y^2+2x^2y+xy+1=-\frac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y+1\right)\left(xy+1\right)=-\frac{1}{4}\\\left(x^2+y\right)^2+xy+1=-\frac{1}{4}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy+1=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\a^2+b=-\frac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\b=-\frac{1}{4}-a^2\end{matrix}\right.\)

\(\Rightarrow\left(a+1\right)\left(-\frac{1}{4}-a^2\right)=-\frac{1}{4}\)

\(\Leftrightarrow4a^3+4a^2+a=0\Leftrightarrow a\left(2a+1\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\Rightarrow b=-\frac{1}{4}\\a=-\frac{1}{2}\Rightarrow b=-\frac{1}{2}\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x^2+y=0\\xy+1=-\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-x^2\\-x^3=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)

TH2: \(\left\{{}\begin{matrix}x^2+y=-\frac{1}{2}\\xy+1=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-\frac{1}{2}-x^2\\x\left(-\frac{1}{2}-x^2\right)=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)

b/ ĐKXĐ; ...

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3x^2+3x+1-16x-16=\frac{8}{y^3}-\frac{8}{y}\\5\left(x^2+2x+2\right)=1+\frac{4}{y^2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\frac{8}{y^3}-\frac{8}{y}\\5\left(x+1\right)^2=\frac{4}{y^2}-4\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=a\\\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3-16a=8b^3-8b\\5a^2=4b^2-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^3-8b^3=16a-8b\\4=-5a^2+4b^2\end{matrix}\right.\)

Nhân vế với vế:

\(4\left(a^3-8b^3\right)=4\left(4a-2b\right)\left(-5a^2+4b^2\right)\)

\(\Leftrightarrow21a^3-10a^2b-16ab^2=0\)

\(\Leftrightarrow a\left(21a^2-10ab-16b^2\right)=0\)

\(\Leftrightarrow a\left(7a-8b\right)\left(3a+2b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\7a=8b\\3a=-2b\end{matrix}\right.\) \(\Rightarrow...\)

\(\frac{a}{b}=\frac{-3}{4}\Rightarrow a=-3k;b=4k\Rightarrow a+5b=17k=34\Rightarrow k=2\Rightarrow a=-6;b=8\) 

Quân đây nhé 

a) \(\frac{3x-2}{x+1}=\frac{6x-4}{2x+2}=\frac{6x-10}{2x+8}=\frac{6x-4-6x+10}{2x+2-2x-8}=\frac{6}{-6}=-1\)

\(\Rightarrow\)\(3x-2=-x-1\)\(\Leftrightarrow\)\(x=\frac{1}{4}\)

b) \(\frac{x}{y}=\frac{-3}{y}\)\(\Leftrightarrow\)\(\frac{x}{-3}=\frac{y}{4}\)\(\Leftrightarrow\)\(\frac{x}{-3}=\frac{5y}{20}=\frac{x+5y}{-3+20}=\frac{34}{17}=2\)

\(\Rightarrow\)\(\hept{\begin{cases}x=2.\left(-3\right)=-6\\y=2.4=8\end{cases}}\)

a. \(\frac{3x-2}{x+1}=\frac{6x-10}{2x+8}\)

<=> (3x - 2)(2x + 8) = (6x - 10)(x + 1)

<=> 6x² - 4x + 24x - 16 = 6x² - 10x + 6x - 10

<=> 6x² + 20x - 16 = 6x² - 4x - 10

<=> 24x = 6

<=> x = \(\frac{1}{4}\)

b. \(\frac{x}{y}=\frac{-3}{4}\) => \(\frac{x}{-3}=\frac{y}{4}\)

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

\(\frac{x}{-3}=\frac{y}{4}=\frac{x+5y}{-3+20}=\frac{34}{17}\)= 2

=> x = 2(-3) = -6

     y = 2.4 = 8

Vậy ...