a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

\(\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)

b) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(=\left(x-y+z\right)^2+\left(y-z\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(=\left(x-y+z+y-z\right)^2=x^2\)

a) \(\left(x-y+x+y\right)^2=4x^2\)

b) \(\left(x-y+z+y-z\right)^2=x^2\)

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

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x+2y}{7+2\cdot5}=\dfrac{51}{17}=3\)

Do đó: x=21; y=15

a) \(\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{2y}{10}=\dfrac{x+2y}{7+10}=\dfrac{51}{17}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.7=21\\y=3.5=15\end{matrix}\right.\)

b) \(\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{9}\\y^2=\dfrac{16}{9}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{4}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)

c) \(\dfrac{x}{y}=\dfrac{2}{5}\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

\(\Rightarrow xy=10k^2=40\Rightarrow k=\pm2\)

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

\(a,5x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x+2y}{7+10}=\dfrac{51}{17}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x=21\\y=15\end{matrix}\right.\)

\(b,\) Áp dụng t/c dtsbn:

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{9}\\y=\dfrac{4}{9}\end{matrix}\right.\)

\(c,\dfrac{x}{y}=\dfrac{2}{5}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\Leftrightarrow x=2k;y=5k\)

\(xy=40\Leftrightarrow10k^2=40\Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4;y=10\\x=-4;y=-10\end{matrix}\right.\)

Bài 1:

Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{x+y}{2+4}=\frac{-12}{6}=-2\)

\(\Rightarrow x=-4,y=-8,z=-10\)

Vậy \(x=-4,y=-8,z=-10\)

Bài 2:

Giải:

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

\(\frac{x}{3}=\frac{y}{4}=\frac{2y}{8}=\frac{2y-x}{8-3}=\frac{10}{5}=2\)

\(\Rightarrow x=6,y=8\)

Vậy \(x=6,y=8\)

1. Từ x/2=y/4=z/5 và x+y=-12

=>x/2=y/4=x+y/2+4=-12/6=-2

=>x/2=-2=>x=-4

=>y/4=-2=>y=-8

=>z/5=-2=>z=-10

Vậy x=-4;y=-8;z=-10

2.Từ x/3=y/4 và 2y-x=10

=>x/3=y/4=2y/8=2y-x/8-3=10/5=2

=>x/3=2=>x=6

=>y/4=2=>y=8

Vậy x=6;y=8

a/ theo bài ra, ta có:

\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z\)

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

\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+z+x+1+x+y-2}=\frac{x+y+z}{2\left(x+y+z\right)}=x+y+z\)

=> y + z = \(\frac{1}{2}\) - x

=> z + x = \(\frac{1}{2}\) - y

=> x + y = \(\frac{1}{2}\) - z

=> \(\frac{x}{\frac{1}{2}-x+1}=\frac{y}{\frac{1}{2}-y+1}=\frac{z}{\frac{1}{2}-z-2}=\frac{1}{2}\)

=> 2x = \(\frac{1}{2}\) - x + 1 => x = \(\frac{1}{2}\)

=> 2y = \(\frac{1}{2}-y+1\) => y = \(\frac{1}{2}\)

=> 2z = \(\frac{1}{2}-z-2\) => z = \(\frac{-1}{2}\)

vậy x = 0 hoặc 1/2

y = 0 hoặc 1/2

z = 0 hoặc -1/2

mk lm câu b bái 1 nha

Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\\=\frac{2x+3y-z-2-6+3}{9}=\frac{2x+3y-z-5}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)

Suy ra

x - 1 = 5 . 2 = 10

x = 10 + 1

→ x = 11

y - 2 = 3 . 5 = 15

y = 15 + 2

→ y = 17

z - 3 = 4 . 5 = 20

z = 20 + 3

→ z = 23

c/ Theo bài ra, ta có:

10x = 6y => \(\frac{x}{6}=\frac{y}{10}\)

=> \(\left(\frac{x}{6}\right)^2=\left(\frac{y}{10}\right)^2=\frac{x^2}{36}=\frac{y^2}{100}=\frac{2x^2}{72}\)

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

=> \(\frac{2x^2}{72}=\frac{y^2}{100}=\frac{2x^2-y^2}{72-100}=\frac{-28}{-28}=1\)

=> 2x²= 72 => x = 6

=> y² = 100 => y = 10

vậy x = 6, y = 10

\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)

= \(\left[\left(x+y+z\right)-\left(x+y\right)\right]^2\)

= \(z^2\)

Ta có:(x + y + z)² - 2(x + y + z) (x + y) + (x + y)²

=[(x+y+z)-(x+y)]²=z²

a, x.(x-y) +y.(x+y)

=x²-xy+xy+y²

=x²+y²

b, (x²-5).(2x+3)-2x.(x-3)

=2x³+3x²-10x-15-2x²+6x

=2x³-x²-4x-15

c, 8-5x.(x+2) +4 .( x-2) . (x+1) +2.( x+2)+ 2.(x-2)+10

=8-5x²-10x+4.(x²+x-2x-2)+2x+4+2x-4+10

=18-6x-5x²+4x²+4x-8x-8

=10-10x-x²

a)  x.(x-y) +y.(x+y)

= x^2-xy+xy+y^2

=x^2+y^2

b) (x^2-5).(2x+3)-2x(x-3)

= 2x^3+3x^2-10x-15-2x^2+6x

=2x^3+x^2-4x-15