Ta có: 4x = 7y

<=> \(\dfrac{4}{y}=\dfrac{7}{x}\)

<=> \(\dfrac{16}{y^2}=\dfrac{49}{x^2}\)

=> \(\dfrac{16+49}{x^2+y^2}=\dfrac{65}{260}=\dfrac{1}{4}\)

=> \(\left\{{}\begin{matrix}x=28\\y=16\end{matrix}\right.\)

Lời giải:

$4x=7y\Rightarrow \frac{x}{7}=\frac{y}{4}=k$

$\Rightarrow x=7k; y=4k$

Ta có: $x^2+y^2=260$

$\Leftrightarrow (7k)^2+(4k)^2=260$

$\Leftrightarrow  65k^2=260$

$\Leftrightarrow k^2=4$

$\Rightarrow k=\pm 2$
Nếu $k=2$ thì $x=7k=14; y=4k=8$. Ta có $(x,y)=(14,8)$

Nếu $k=-2$ thì $x=7k=-14; y=4k=-8$. Ta có $(x,y)=(-14,-8)$

Đặt x=2k; y=4k

x^2.y^2=4 

(2k)^2.(4k)^2=4

64.k^4=4

k^4=1/16

k=1/2 hoặc k=-1/2

Khi k=1/2 --> x=2.1/2=1

y=4.1/2=2

Khi k=-1/2 --> x=2.(-1/2)=-1

y=4.(-1/2)=-2

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

⇔\(\left(x+4\right)\left(x+4\right)=100\)

⇔\(\left(x+4\right)^2=10^2\)

⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

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

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

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

⇒\(\left\{{}\begin{matrix}x=5.15=75\\y=5.10=50\\z=5.6=30\end{matrix}\right.\)

\(A=\dfrac{1}{2}\left(x-3\right)^2+10\ge10\\ A_{min}=10\Leftrightarrow x-3=0\Leftrightarrow x=3\)

\(A=\dfrac{1}{2}\left(x-3\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi x=3

\(x:y:z=3:5:\left(-2\right)\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=-\dfrac{16}{4}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-4\right).3=-12\\y=\left(-4\right).5=-20\\z=\left(-4\right).\left(-2\right)=8\end{matrix}\right.\)

x:y:z=4:5:6

--> x/4=y/5=z/6

Đặt x=4k; y=5k; z=6k

x^2-2y^2+z^2=18

(4k)^2-2.(5k)^2+(6k)^2=18

2k^2=18

k^2=9

k=3 hoặc k=-3

Khi k=3

--> x=4.3=12

y=5.3=15

z=6.3=18

Khi k=-3

--> x=4.(-3)=-12

y=5.(-3)=-15

z=6.(-3)=-18

a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

⇔\(7\left(x-3\right)=5\left(x+5\right)\)

⇔\(7x-21=5x+25\)

⇔\(7x-21-5x-25=0\)

⇔\(2x-46=0\)

⇔\(2x=46\)

⇔\(x=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

⇔\(\left(x+1\right)\left(x-1\right)=7.9\)

⇔\(x^2-1=63\)

⇔\(x^2=64=8^2\)

⇔\(\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(a.\dfrac{x-3}{x+5}=\dfrac{5}{7}\)                          

    \(\left(x-3\right).7=\left(x+5\right).5\) 

    \(\left(x.7\right)+\left[\left(-3\right).7\right]=\left(x.5\right)+\left(5.5\right)\)

    \(7x-21=5x+25\)

    \(7x-5x=25+21\)

             \(2x=46\)

               \(x=46:2\)

               \(x=23\)

Câu b/ cứ làm theo câu a/ là được

AcCl₃ và NiCl₂

a) Áp dụng t/c dtsbn:

 \(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)

b) \(\dfrac{3}{x}=\dfrac{7}{y}\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)

Và \(x+16=y\Rightarrow y-x=16\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{y-x}{7-3}=\dfrac{16}{4}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.3=12\\y=4.7=28\end{matrix}\right.\)

a) \(\dfrac{x}{7}=\dfrac{y}{13}\)

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

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

⇒\(\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)