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28 tháng 9 2020

a) Đề chắc là: \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\)

Ta có: \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\left(\forall x,y,z\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{378}{395}\\z=2004\end{cases}}\)

28 tháng 9 2020

b) Ta có: \(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\left(\forall x,y,z\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|x+\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}\)

1 tháng 8 2017

a,

\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)

d,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)

1 tháng 8 2017

Bạn mới hỏi ở dưới rồi :v

9 tháng 6 2017

Ta có : |3x - 4| + |3y + 5| = 0 

Mà : \(\left|3x-4\right|\le0\forall x\in R\)

         \(\left|3y+5\right|\ge0\forall x\in R\)

Nên |3x - 4| = |3y + 5| = 0 

Suy ra : 3x - 4 = 0 ; 3y + 5 = 0

    =>     3x = 4 ; 3y = -5

    => x = 4/3 ; y = -5/3

1 tháng 8 2017

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)

mà A\(\le0\)

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\)​ phải bằng 0 đê thỏa mãn điều kiện

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy....

b;c)I hệt câu a nên làm tương tự nhá

d)

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=B\)

B=\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\)

Thay ra ta tính đc :\(z=-\dfrac{11}{20}\)

Vậy....

1 tháng 8 2017

thanks bn nha

23 tháng 6 2017

Vì \(\left|x+\frac{19}{5}\right|\ge0\) với \(\forall x\)

\(\left|y+\frac{1890}{1975}\right|\ge0\) với \(\forall y\)

\(\left|z-2004\right|\ge0\)với \(\forall z\)

\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2004\end{cases}}\)

4 tháng 7 2018

a) \(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)

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

\(2\frac{1}{3}-\frac{3}{2}=3x-\frac{3}{2}x-x\)

\(\frac{5}{6}=3x-\frac{3}{2}x-x\)

\(\frac{5}{6}=\left(3-\frac{3}{2}-1\right)x\)

\(\frac{5}{6}=\frac{1}{2}x\)

\(x=\frac{5}{6}:\frac{1}{2}\)

\(x=\frac{5}{3}\)

b) |3x-4|+|3y+5|=0

ĐK : \(\hept{\begin{cases}\left|3x-4\right|\ge0\\\left|3y+5\right|\ge0\end{cases}}\Leftrightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\)

Mà |3x-4|+|3y+5|=0 nên :

\(\Rightarrow\hept{\begin{cases}3x-4=0\\3y+5=0\end{cases}}\Rightarrow\hept{\begin{cases}3x=4\\3y=-5\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{-5}{3}\end{cases}}\)

Vậy x=4/3 ; y=-5/3

c) \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\)

ĐK : \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{1890}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{cases}}\Leftrightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\)

Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\) nên :

\(\Rightarrow\hept{\begin{cases}x+\frac{19}{5}=0\\y+\frac{1890}{1975}=0\\z-2004=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2004\end{cases}}\)

Vậy ...

21 tháng 9 2016

a) 

Ta có : \(\left|x+\frac{19}{5}\right|\ge0\) với mọi x

           \(\left|y+\frac{1890}{1975}\right|\ge0\) với mọi x

            \(\left|z-2014\right|\ge0\) với mọi x

\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2014\right|\ge0\)

Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2014\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{19}{5}\right|=0\\\left|y+\frac{1890}{1975}\right|=0\\\left|z-2014\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x+\frac{19}{5}=0\\y+\frac{1890}{1975}=0\\z-2014=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2014\end{cases}}\)

 b) Cx tương tự câu trên thôi bạn

Ta có : \(\left|x-\frac{9}{2}\right|\ge0\) với mọi x

            \(\left|y+\frac{4}{3}\right|\ge0\) với mọi x

            \(\left|z+\frac{7}{2}\right|\ge0\) với mọi x

\(\Rightarrow\left|x-\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\) với mọi x

Mà \(\left|x-\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)

\(\Rightarrow\hept{\begin{cases}\left|x-\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x-\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}\)