Gợi ý nhá

Bài 3: câu 1: làm tương tự như câu hỏi lần trước bạn gửi.

b)  Bạn chỉ cần cho tử và mẫu mũ 3 lên.  theé là dễ r

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow=\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow=\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

tự tính tiếp =)

1)a)

x/3=y/4=>x/15=y/20

y/5=z/7=>y/20=z/28

=>x/15=y/20=z/18

Áp dụng tính chất dãy tỉ số bằng nhau ta có

x/15=y/20=z/28=2x+3y-z/30+60-28=372/62=6

=>x=90

y=120

z=168

b)

2x=3y=5z

2x=3y=>x/3=y/2=>x/15=y/10

3y=5z=>y/5=z/3=>y/10=z/6

Áp dụng tính chất dãy tỉ số bằng nhau ta có

x/15=y/10=z/6=x+y-z/15+10-6=95/19=5

=>x=75

y=50

z=30

a) Ta co :x/3=y/4 suy ra x/15=y/20 (1)

y/5=z/7 suy ra y/20=z/28 (2)

Tu (1) va (2) suy ra y/20=x/15=z/28

còn lại tự làm nhé dễ rùi

b)Ta co : 2x=3y=5z suy ra x phan 1/2=y phan 1/3 = z phan 1/5

de rui tu lam nha 

a,Ta có : \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\)

\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{10}=\frac{z}{8}\)

Suy ra :\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=k\Rightarrow x-15k;y=10k;z=8k\)

Ta có : \(4(15k)-3(10k)+5(8k)=7\)

\(\Rightarrow60k-30k+40k=7\)

\(\Rightarrow70k=7\). Suy ra \(k=\frac{1}{10}\)

Ta có : \(x=\frac{1}{10}\cdot15=\frac{3}{2}\)

\(y=\frac{1}{10}\cdot10=1\)

Mình chỉ giải có chừng này thôi

Câu b mk làm sau

\(xy+2x-y=7\)

\(xy+2x=7+y\)

\(x\left(y+2\right)=7+y\)

\(x=\frac{7+y}{y+2}\)

a) Vì \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\)

        \(3y=7z\Rightarrow\frac{y}{7}=\frac{z}{3}\Rightarrow\frac{y}{14}=\frac{z}{6}\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{6}\) và x+y-z=58

APa dụng TC dãy TSBN ta có

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{6}=\frac{x+y-z}{21+14-6}=\frac{58}{29}=2\)

\(\Rightarrow x=42;y=28;z=12\)

Các câu còn lại tương tự

cách 2: 2x + 1 = 3y + 3= 5z + 3 => \(2.\left(x+\frac{1}{2}\right)=3.\left(y+1\right)=5.\left(z+\frac{3}{5}\right)\)

=> \(\frac{2.\left(x+\frac{1}{2}\right)}{30}=\frac{3\left(y+1\right)}{30}=\frac{5\left(z+\frac{3}{5}\right)}{30}\) => \(\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z+\frac{3}{5}}{6}\)

Áp dụng tc dãy tỉ số bằng nhau => \(\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z+\frac{3}{5}}{6}=\frac{x+\frac{1}{2}-y-1+z+\frac{3}{5}}{15-10+6}=\frac{1,1+\frac{1}{10}}{11}=\frac{6}{55}\)

=> \(x+\frac{1}{2}=\frac{6}{55}.15=\frac{18}{11}\Rightarrow x=\frac{25}{11}\)

tương tự, y = ...; z  ...

có cách khác k 

bảo t vs t sắp hok rùi

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{38}{19}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.15=30\\y=2.10=20\\z=2.6=12\end{matrix}\right.\)

\(2x=3y=5z\) \(\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Theo tính chất dãy tỉ số bằng nhau: 

   \(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{38}{19}=2\)

  \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=2\\\dfrac{y}{10}=2\\\dfrac{z}{6}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=30\\y=20\\z=12\end{matrix}\right.\)

\(2x=3y-2x\Leftrightarrow4x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\\ 3y-2x=5z\Leftrightarrow4x-2x=5z\Leftrightarrow2x=5z\Leftrightarrow\dfrac{x}{5}=\dfrac{z}{2}\\ \Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{6}=\dfrac{x-y+z}{15-20+6}=\dfrac{99}{1}=99\\ \Leftrightarrow\left\{{}\begin{matrix}x=1485\\y=1980\\z=594\end{matrix}\right.\)