Question

[Something Just Like This]

1. Tìm x ; y ; z biết :

a) \(2009-\left|x-2009\right|=x\)

b) \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

2. Tìm a,b,c biết :

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\) và a+b+c=-50

Answer 1

[Hoàng Thị Ngọc Anh]

1. a) \(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=2009-x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy \(x\le2009.\)

b) Ta có: \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}\ge0\forall x\\\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\\\left|x+y-z\right|\ge0\forall x,y,z\end{matrix}\right.\) \(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\forall x,y,z\)

Dấu \("="\) xảy ra khi \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}=0\\\left(y-\dfrac{2}{5}\right)^{2008}=0\\\left|x+y-z\right|=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\).

Answer 2

[Nguyễn Nguyệt Hằng]

Bạn kia làm câu 1 rồi thì mình làm câu 2 nhé!

2. Ta có:\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\)

\(\Rightarrow\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{5b-3c}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{15a-10b+6c-15a}{25+9}\)=\(\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)

\(\Rightarrow\dfrac{-5b+3c}{17}=\dfrac{5b-3c}{2}\Rightarrow5b-3c=0\)

=> 5b=3c =>\(\left\{{}\begin{matrix}b=\dfrac{3}{5}c\\a=\dfrac{2}{5}c\end{matrix}\right.\)

=>\(\dfrac{3}{5}c+\dfrac{2}{5}c+c=-50\)

=> \(c\left(\dfrac{3}{5}+\dfrac{2}{5}+1\right)=-50\)

=> 2c = -50

=> c= -25

=>\(\left\{{}\begin{matrix}b=-25.\dfrac{3}{5}=-15\\a=-25.\dfrac{2}{5}=-10\end{matrix}\right.\)

Vậy a= -10; b= -15; c= -25