a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)

\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)

\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)

Đến đây tự làm tiếp nhé

b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

=> x = 75, y = 50, z = 30

c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)

\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)

\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)

=> x=... , y=... , z=...

d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)

Ta có: xy = 90 => 2k.5k = 90 => 10k² = 90 => k² = 9 => k = 3 hoặc -3

Với k = 3 => x = 6, y = 15

Với k = -3 => x = -6, y = -15

Vậy...

e, Tương tự câu d

b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)

=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)

     \(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)

      \(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)

b) \(\dfrac{4}{14}-\dfrac{1}{5}+\dfrac{11}{15}-\dfrac{24}{5}\)

\(=-\dfrac{1}{5}-\dfrac{24}{5}+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-\left(\dfrac{1}{5}+\dfrac{24}{5}\right)+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-\left(\dfrac{1+24}{5}\right)+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-\dfrac{25}{5}+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-5+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=\dfrac{-70}{14}+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=\dfrac{-70+4}{14}+\dfrac{11}{15}\)

\(=\dfrac{-\left(70-4\right)}{14}+\dfrac{11}{15}\)

\(=\dfrac{-66}{14}+\dfrac{11}{15}\)

\(=\dfrac{-990}{210}+\dfrac{154}{210}\)

\(=\dfrac{-990+154}{210}\)

\(=\dfrac{-\left(990-154\right)}{210}\)

\(=\dfrac{-836}{210}\)

\(=\dfrac{-418}{105}\)

-Đang xy sao thành ab vậy bạn?

a) Ta có : \(x - 2xy + y - 3 = 0\)

\(\Rightarrow-2xy+x+y=3\)

\(\Rightarrow-2.\left(-2xy+x+y\right)=-2.3\)

\(\Rightarrow4xy-2x-2y=-6\)

\(\Rightarrow4xy-2x-2y+1=-6+1\)

\(\Rightarrow2x.\left(2y-1\right).\left(2y-1\right)=-5\)

\(\Rightarrow\left(2y-1\right).\left(2x-1\right)=-5=1.\left(-5\right)=-5.1=\left(-1\right).5=5.\left(-1\right)\)

Tự lập bảng đi -.-

Vậy có 5 bộ số (x, y, z) thoã mãn: (0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)

❏Ta có : \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{2}\\xy=54\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\xy=54\end{matrix}\right.\)

\(\Rightarrow x^2=54\Rightarrow x=y=\sqrt{54}=3\sqrt{6}\)

Ta có : \(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{2}\\xy=54\end{matrix}\right.=>\left\{{}\begin{matrix}x=y\\xy=54\end{matrix}\right.\)

⇒\(x^2=54\)=>x=y=\(\sqrt{54=3\sqrt{6}}\)

a.

\(\frac{2x}{7}=\frac{3y}{2}\Rightarrow4x=21y\)

\(x-y=17\Rightarrow x=17+y\)

\(\Rightarrow4\left(17+y\right)=21y\Rightarrow68+4y=21y\Rightarrow17y=68\Rightarrow y=4\)

\(\Rightarrow x=17+y=17+4=21\)

b.

\(\frac{x}{2}=\frac{y}{5}\Rightarrow5x=2y\)

\(x\cdot y=40\Rightarrow x=\frac{40}{y}\)

\(\Rightarrow5\cdot\frac{40}{y}=2y\Rightarrow\frac{200}{y}=2y\Rightarrow2y^2=200\Rightarrow y=\pm10\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

\(b.\)

Theo đề : \(2x=3y=5z\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\) và \(x+y-x=95\)

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

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

\(\Rightarrow x=75;y=50;z=30\)

\(d.\)

Đặt : \(\dfrac{x}{2}=\dfrac{y}{5}=k\)

\(\Rightarrow x=2k;y=5k\)

Thay \(x=2k;y=5k\) vào \(xy=90\)

\(\left(2k\right)\left(5k\right)=90\)

\(\Rightarrow10k^2=90\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3\)

+ Nếu \(k=3\Rightarrow x=6;y=15\)

+ Nếu \(k=-3\Rightarrow x=-6;y=-15\)

\(e.\)

Tương tự với câu \(d\)