Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\Leftrightarrow x=15k;y=20k;z=40k\)

\(xy=1200\\ \Leftrightarrow300k^2=1200\\ \Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30;y=40;z=80\\x=-30;y=-40;z=-80\end{matrix}\right.\)

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

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

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

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

\(\sum\sqrt{\dfrac{1+x^3+y^3}{xy}}\ge\sum\sqrt{\dfrac{3xy}{xy}}\ge3\sqrt{3}\)

chắc là bạn ghi sai đề rồi -_- ;

\(P=\dfrac{x+2y}{2xy}+\dfrac{1}{x+2y}=\dfrac{x+2y}{4}+\dfrac{1}{x+2y}\)

\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)

\(P\ge2\sqrt{\dfrac{x+2y}{16\left(x+2y\right)}}+\dfrac{3}{16}.2\sqrt{2xy}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)

a) Ta có :\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)

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

\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)\(=\dfrac{5x+2y+z}{555+444+333}=\dfrac{1100}{1332}=\dfrac{275}{333}\)

Từ đó tìm được x;y;z

b) Từ \(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)

Đặt \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4k\\y^2=9k\end{matrix}\right.\)

\(\Rightarrow x^2\cdot y^2=4k\cdot9k=52\)

\(\Rightarrow36k^2=52\)

\(\Rightarrow k^2=\dfrac{13}{9}\) (sai đề)

b: Sửa đề: x^2+y^2=52

Đặt x/2=y/3=k

=>x=2k; y=3k

x^2+y^2=52

=>4k^2+9k^2=52

=>k^2=4

TH1: k=2

=>x=4; y=6

TH2: k=-2

=>x=-4; y=-6

c: Đặt x/5=y/3=k

=>x=5k; y=3k

x^2-y^2=16

=>25k^2-9k^2=16

=>k^2=1

TH1: k=1

=>x=5; y=3

TH2: k=-1

=>x=-5; y=-3

d: Đặt x/2=y/3=k

=>x=2k; y=3k

Ta có: xy=54

=>2k*3k=54

=>6k^2=54

=>k^2=9

TH1: k=3

=>x=6; y=9

TH2: k=-3

=>x=-6; y=-9

e: Đặt x/4=y/3=k

=>x=4k; y=3k

Ta có: xy=12

=>4k*3k=12

=>k^2=1

TH1: k=1

=>x=4; y=3

TH2: k=-1

=>x=-4; y=-3