ta có: \(\frac{x}{\frac{y}{z}}=\frac{3}{\frac{4}{5}}\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}.\)

ADTCDTSBN

...

bn tu lm tiep nha

mà bạn nguyen nguyen ơi, x/3 phải bằng x^2/3*x chứ

+) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

=> \(\hept{\begin{cases}\frac{x^2}{9}=4\\\frac{y^2}{16}=4\end{cases}}\) => \(\hept{\begin{cases}x^2=4.9=36\\y^2=4.16=64\end{cases}}\) => \(\hept{\begin{cases}x=\pm6\\y=\pm8\end{cases}}\)

Vậy ...

`x : y : z= 3:4:5`

`=> x/3 = y/4 = z/5 <=> x^2/9 = y^2/16 = z^2/25`

Áp dụng dãy tỉ số bằng nhau:

`x^2/9 = y^2/16 = z^2/25 = (2x^2 + 2y^2 - 3z^2)/(18 + 32 - 75) = -100/-25 = 4`.

`=> {(x^2/9 = 4 => x = +-6), (y^2/16 =4 <=> x = +-8), (z^2/25 = 4 => z = +-10):}`

Vậy ...

\(x-\frac{1}{2}=y-\frac{2}{3}=z-\frac{3}{4}\)va \(x-2y+3z=14\)

\(\frac{\Rightarrow\left(x-1\right)}{2}=\frac{\left(-2y+4\right)}{-6}=\frac{\left(3z-9\right)}{12}\)

\(=\frac{\left(x-1-2y+4+3z-9\right)}{\left(2-6+12\right)}\)

\(\Rightarrow-\frac{16}{8}=-2\)

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

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

\(\Rightarrow\frac{\left(x-3\right)}{4}=-2\Leftrightarrow z-3=-8\Leftrightarrow z=-5\)

\(b)\)

Theo đề ra:

\(x:y:z=3:4:5\)

\(2x^2+2y^2-3z^2=-100\)

\(\Leftrightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

\(\Leftrightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)

\(\Leftrightarrow\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}\)

Áp dụng tính chất dãy tỷ số bằng nhau:

\(\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x}{3}=4\Leftrightarrow x=12\\\frac{y}{4}=4\Leftrightarrow y=16\\\frac{z}{5}=4\Leftrightarrow z=20\end{cases}}\)

đừng nên dựa vào trang này quá 

bài trên thuộc dạng SGK , SBT mà không làm được à

a, Theo đề bài ta có :

x : y : z = 3 : 4 : 5

=>\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\)

ADTCDTSBN:

\(\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}=\dfrac{2x^2+2y^2-3z^2}{18+32+75}=\dfrac{-4}{5}\)

\(\dfrac{x}{3}=\dfrac{-4}{5}\Rightarrow x=\dfrac{-12}{5}\)

\(\dfrac{y}{4}=\dfrac{-4}{5}\Rightarrow y=\dfrac{-16}{5}\)

\(\dfrac{z}{5}=\dfrac{-4}{5}\Rightarrow z=-4\)

\(x:y:z=3:4:5=>\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

\(=>x=\dfrac{3y}{4},z=\dfrac{5y}{4}\) thay x,z vào \(2x^2+2y^2-3z^2=-100\)

\(< =>2\left(\dfrac{3y}{4}\right)^2+2y^2-3\left(\dfrac{5y}{4}\right)^2=-100\)

\(=>y=\pm8\)

* với y=8 \(=>x=\dfrac{3.8}{4}=6,z=\dfrac{5.8}{4}=10\)

* với y=-8 \(=>x=-6,z=-10\)

hinh nhu vao google tim cug dc ban co the kham khao rat nhieu bai

x : y : z = 3 : 4 : 5

=>x/3=y/4=z/5 => x²/9=y²/16=z²/5 = 2x²=2x²/18=2y²/32=3z²/75

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)

suy ra 2x²/18=4 =>x²=36 =>x=6 ; x=-6

2y²/32=4 =>x²=128 => y=8 ; y=-8

3x²/75=4 =>z²=100 =>z=10 ;z=-10

a) Đặt: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\) 

Ta có: \(x^2+3y^2-2z^2=-16\)

\(\Rightarrow\left(2k\right)^2+3\cdot\left(3k\right)^2-2\cdot\left(4k\right)^2=-16\)

\(\Rightarrow4k^2+3\cdot9k^2-2\cdot16k^2=-16\)

\(\Rightarrow4k^2+27k^2-32k^2=-16\)

\(\Rightarrow-k^2=-16\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k=\pm4\)

Với k = 4

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{y}{3}=4\\\dfrac{z}{4}=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot4=8\\y=3\cdot4=12\\z=4\cdot4=16\end{matrix}\right.\)

Với k = -4 

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-4\\\dfrac{y}{3}=-4\\\dfrac{z}{4}=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot-4=-8\\y=3\cdot-4=-12\\z=4\cdot-4=-16\end{matrix}\right.\) 

Vậy: ...

b) Đặt: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) 

Ta có: \(2x^2+2y^2-3z^2=-100\)

\(\Rightarrow2\cdot\left(3k\right)^2+2\cdot\left(4k\right)^2-3\cdot\left(5k\right)^2=-100\)

\(\Rightarrow2\cdot9k^2+2\cdot16k^2-3\cdot25k^2=-100\)

\(\Rightarrow18k^2+32k^2-75k^2=-100\)

\(\Rightarrow-25k^2=-100\)

\(\Rightarrow k^2=-\dfrac{100}{-25}=4\)

\(\Rightarrow k=\pm2\)

Với k = 2

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=2\\\dfrac{y}{4}=2\\\dfrac{z}{5}=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=2\cdot4=8\\z=2\cdot5=10\end{matrix}\right.\)

Với k = -2

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-2\\\dfrac{y}{4}=-2\\\dfrac{z}{5}=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot-3=-6\\y=2\cdot-4=-8\\z=2\cdot-5=-10\end{matrix}\right.\)

Vậy: ... 

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{2x^2+2y^2-3z^2}{2\cdot9+2\cdot16-3\cdot25}=\frac{-100}{-25}=4\)

\(\Rightarrow x^2=36;y^2=64;z^2=100\)

\(\Rightarrow\) x = + 6; y = + 8; z = + 10