dùng dãy tỉ số bàng nhau ấy ngu vậy

không nên nói tục

a, \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{3}=\frac{5y}{10}\)

Áp dụng tính chất Dãy tỉ số bằng nhau

\(\frac{x}{3}=\frac{5y}{10}=\frac{x-5y}{3-10}=\frac{2,1}{-7}\frac{-3}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{3}=-\frac{3}{10}\\\frac{y}{2}=-\frac{3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{9}{10}\\y=-\frac{3}{5}\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=-\frac{9}{10}\\y=-\frac{3}{5}\end{matrix}\right.\)

b, Đặt \(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}=k\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{3}=k\\\frac{y}{7}=k\\\frac{z}{2}=k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\\z=2k\end{matrix}\right.\)

2x²+y²+3z²= 2.(3k)²+(7k)²+3.(2k)²

316= 18k²+49k²+12k²

316=k².(18+49+12)

316=79k²

4=k²

\(\Rightarrow\left[{}\begin{matrix}k=-2\\k=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3k\\y=7k\\z=2k\end{matrix}\right.\\\left\{{}\begin{matrix}x=3k\\y=7k\\z=2k\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-6\\y=-14\\z=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\y=14\\z=4\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=-6\\y=-14\\z=-4\end{matrix}\right.\)

và \(\left\{{}\begin{matrix}x=6\\y=14\\z=4\end{matrix}\right.\)

*Chúc bạn học tốt*

Chiều nay mk phải nộp rồi giúp mk với !

Bài giải

a, \(2x=3y\text{ }\Rightarrow\text{ }\frac{x}{3}=\frac{y}{2}=\frac{5y}{10}=\frac{x-5y}{3-10}=\frac{2,1}{-7}=\frac{-3}{10}\)

( Áp dụng t/c ... )

\(\Rightarrow\text{ }x=\frac{-3}{10}\cdot3=-\frac{9}{10}\)

\(y=\frac{-3}{10}\cdot2=-\frac{6}{10}\)

b, \(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}\Rightarrow\text{ }\frac{x^2}{9}=\frac{y^2}{49}=\frac{z^2}{4}\text{ }=\frac{2x^2}{18}=\frac{3z^2}{12}=\frac{2x^2+y^2+3z^2}{18+49+12}=\frac{316}{79}=4\)

\(\Rightarrow\text{ }x^2=4\cdot9=36\text{ }\Rightarrow\text{ }x=\pm6\)

\(y^2=4\cdot49=196\text{ }\Rightarrow\text{ }y=\pm14\)

\(z^2==4\cdot4=16\text{ }\Rightarrow\text{ }z=4\)

A= x^2 - 2x(y+7) + (y+7)^2 -(y+7)^2 + 6y^2 - 6y +72

=(x-y-7)^2 + 5(y^2 - 4y +4) +101

=(x-y-7)^2 + 5(y-2)^2 +101\(\ge\)101

\(\Rightarrow\)Min A= 101\(\Leftrightarrow\)x=9;y=2
 

Đề?

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{2}\)

\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{49}=\dfrac{z^2}{4}\)

\(\Rightarrow\dfrac{2x^2}{18}=\dfrac{y^2}{49}=\dfrac{3z^2}{12}\)

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

\(\dfrac{2x^2}{18}=\dfrac{y^2}{49}=\dfrac{3z^2}{12}=\dfrac{2x^2+y^2+3z^2}{18+49+12}=\dfrac{316}{79}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.18:2=36\\y^2=4.49=196\\z^2=4.12:3=16\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=6\\y=14\\z=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-14\\z=-4\end{matrix}\right.\end{matrix}\right.\)

x, y, z tỉ lệ với 3, 7, 2

=> \(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}\)

Đặt \(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}=k\Rightarrow\hept{\begin{cases}x=3k\\y=7k\\z=2k\end{cases}}\)

2x² + y² + 3z² = 316

<=> 2.(3k)² + (7k)² + 3.(2k)² = 316

<=> 2.9k² + 49k² + 3.4k² = 316

<=> 18k² + 49k² + 12k² = 316

<=> 79k² = 316

<=> k² = 4

<=> k = ±2

Với k = 2 => \(\hept{\begin{cases}x=3\cdot2=6\\y=7\cdot2=14\\z=2\cdot2=4\end{cases}}\)

Với k = -2 => \(\hept{\begin{cases}x=3\cdot\left(-2\right)=-6\\y=7\cdot\left(-2\right)=-14\\z=2\cdot\left(-2\right)=-4\end{cases}}\)

Vậy ( x ; y ; z ) = { 6 ; 14 ; 4 ) , ( -6 ; -14 ; -4 ) }

Theo bài ra ta có : \(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}\)

Đặt \(\hept{\begin{cases}x=3k\\y=7k\\z=2k\end{cases}}\)Ta có : \(2x^2+y^2+3z^2=316\)

\(2.\left(3k\right)^2+\left(7k\right)^2+3.\left(2z\right)^2=316\)

\(\Leftrightarrow18k^2+49k^2+12k^2=316\Leftrightarrow49k^2=316\Leftrightarrow k=\pm2\)

Tự thay nhé