Tìm x, y, z

\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)

Áp dụng tích chất của dãy tỉ số bằng nhau, ta có

\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)

+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)

+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)

\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)

Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)

Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!

Rút gọn các biểu thức:

a) {\(\dfrac{1}{x^2}\) + \(\dfrac{1}{y^2}\) + \(\dfrac{2}{x+y}\)(\(\dfrac{1}{x}\) + \(\dfrac{1}{y}\))} : \(\dfrac{x^3+y^3}{x^2y^2}\)

b) {\(\dfrac{1}{\left(2x-y\right)^2}\) + \(\dfrac{2}{4x^2-y^2}\) + \(\dfrac{1}{\left(2x+y\right)^2}\)} . \(\dfrac{4x^2+4xy+y^2}{16x}\)

c) (\(\dfrac{x^2-xy}{x^2y+y^3}\) - \(\dfrac{2x^2}{y^3-xy^2+x^2y-x^3}\))(1 - \(\dfrac{y-1}{x}\) - \(\dfrac{y}{x^2}\))

Đặt $ X = a - b; Y = b - c; Z = c - a \Rightarrow X + Y + Z = 0$

Với X + Y + Z = 0, ta chứng minh được :
$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$

Thật vậy, ta có :

$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + \dfrac{2}{XY} + \dfrac{2}{YZ} + \dfrac{2}{ZX}$

$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + 2.\dfrac{X + Y + Z}{XYZ}$

$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$ ( do X + Y + Z = 0)

$ \Rightarrow \sqrt{\dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}} = \sqrt{( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2} = |\dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z}|$

Suy ra : $ \sqrt{\dfrac{1}{(a - b)^2} + \dfrac{1}{(b - c)^2} +\dfrac{1}{( c - a)^2}} = |\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$

Do a, b, c là số hữu tỷ nên $|\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$ cũng là số hữu tỷ. Ta có điều phải chứng minh.

giải các hệ phương trình

a \(\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\)

\(\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\)

b \(\dfrac{5}{x+y-3}-\dfrac{2}{x-y+1}=8\)

\(\dfrac{3}{x+y-3}+\dfrac{1}{x-y+1}=\dfrac{3}{2}\)

c \(\sqrt{x-1}-3\sqrt{y+2}=2\)

\(2\sqrt{x-1}+5\sqrt{y+2}=15\)

d \(\dfrac{7}{\sqrt{x-7}}-\dfrac{4}{\sqrt{y+6}}=\dfrac{5}{3}\)

\(\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\)

e \(7x^2+13y=-39\)

\(5x^2-11y=33\)

f \(2\left(x-1\right)^2-3y^3=7\)

\(5\left(x-1\right)^2+6y^3=4\)