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

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

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

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

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.

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

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

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

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

Đặt k, hoặc lm theo cách này cx đc:

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{xy}{2y}=\dfrac{40}{2y}\)

\(\Rightarrow\dfrac{y}{5}=\dfrac{40}{2y}\Rightarrow2y.y=40.5\)

\(\Rightarrow2y^2=200\Rightarrow y^2=100\)

\(\Rightarrow\left\{{}\begin{matrix}y=10\\y=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{40}{10}=4\\x=\dfrac{40}{-10}=-4\end{matrix}\right.\)

Vậy \(\left(x,y\right)\) là: \(\left(4;10\right);\left(-4;-10\right)\)

Ta có : \(\dfrac{x}{2}=\dfrac{y}{5}=k\Leftrightarrow x=2k;y=5k\)

Mà : x.y=40

\(\Leftrightarrow2k.5k=40\Leftrightarrow10k^2=40\Rightarrow k^2=4\Rightarrow k=\pm2\)

Với \(\dfrac{x}{2}=2\Rightarrow x=4\)

Với \(\dfrac{y}{5}=2\Rightarrow y=10\)

Với \(\dfrac{x}{2}=-2\Rightarrow x=-4\)

Với \(\dfrac{y}{5}=-2\Rightarrow y=-10\)

Vậy x=4 và y=10 hoặc x=-4 và y=-10

\(bx^2=ay^{2^{ }}=\dfrac{x^2}{\dfrac{1}{b}}=\dfrac{y^2}{\dfrac{1}{a}}=\dfrac{x^2+y^2}{\dfrac{a+b}{ab}}=\dfrac{ab}{a+b}.\)

\(\Leftrightarrow\dfrac{x^2}{a}=\dfrac{1}{a+b}=\dfrac{y^2}{b}.\)

\(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}=2.\left(\dfrac{1}{a+b}\right)^{1008}=\dfrac{2}{\left(a +b\right)^{1008}}\left(dpcm\right)\)

Theo bài ra ta có:

\(bx^2=ay^2\) \(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\)

\(x^2+y^2=1\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)

\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{1}{a+b}\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}\) \(=\dfrac{\left(x^2\right)^{1008}}{a^{1008}}+\dfrac{\left(y^2\right)^{1008}}{b^{1008}}\)

\(=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}\)

\(=\left(\dfrac{1}{a+b}\right)^{1008}+\left(\dfrac{1}{a+b}\right)^{1008}\)

\(=2\cdot\left(\dfrac{1}{a+b}\right)^{1008}\)

\(=2\cdot\dfrac{1^{1008}}{\left(a+b\right)^{1008}}\)

\(=2\cdot\dfrac{1}{\left(a+b\right)^{1008}}\)

\(=\dfrac{2}{a+b}^{1008}\)

Vậy \(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{a+b}^{1008}\)

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

\(\text{Câu 1: }\\ \text{Theo bài ra ta có : }x+y-z=10\\ \dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{2}=\dfrac{4y}{12}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\\ \dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{3y}{12}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\\ \text{Từ }\left(1\right)\text{ và }\left(2\right)\text{ suy ra : }\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\\ \text{ Áp dụng tính chất dãy tỉ số bằng nhau ta được : }\\ \dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=16\\\dfrac{y}{12}=2\Rightarrow y=24\\\dfrac{z}{15}=2\Rightarrow z=30\end{matrix}\right.\\ \text{Vậy }x=16\\ y=24\\ z=30\)

\(\text{Câu 2 : }\\ \text{Ta có : }\dfrac{x}{2}=\dfrac{y}{5}\\ \Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{5}\right)^2=\dfrac{x}{2}\cdot\dfrac{y}{5}=\dfrac{xy}{2\cdot5}=\dfrac{7+3}{10}=\dfrac{10}{10}=1\\ \Rightarrow\left\{{}\begin{matrix}\left(\dfrac{x}{2}\right)^2=1\Rightarrow\dfrac{x}{2}=1\Rightarrow x=2\\\left(\dfrac{y}{5}\right)^2=1\Rightarrow\dfrac{y}{5}=1\Rightarrow y=5\end{matrix}\right.\\ \text{Vậy }x=2\\ y=5\)

Câu 3 : \(\dfrac{\text{Giải}}{ }\)

Gọi số học sinh 4 khối \(6,7,8,9\) lần lượt là \(a;b;c;d\) \(\left(a;b;c;d\in N\text{*}\right)\) \(\left(em\right)\)

Theo bài ra ta có : \(b-d=70\)

\(a;b;c;d\) tỉ lệ với \(9;8;7;6\) \(\Rightarrow\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}=\dfrac{b-d}{8-6}=\dfrac{70}{2}=35\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{9}=35\Rightarrow a=315\\\dfrac{b}{8}=35\Rightarrow b=280\\\dfrac{c}{7}=35\Rightarrow c=245\\\dfrac{d}{6}=35\Rightarrow d=210\end{matrix}\right.\)

\(\text{Vậy }a=315\\ b=280\\ c=245\\ d=210\)

a)\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)

\(\dfrac{x}{3}=2\Rightarrow x=6\)

\(\dfrac{y}{7}=2\Rightarrow y=14\)

b)\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{5}=2\Rightarrow x=10\)

\(\dfrac{y}{2}=2\Rightarrow y=4\)