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k cho tui

a,

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

Mà

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

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

d,

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

Mà

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

Bạn mới hỏi ở dưới rồi :v

b,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{2}{5}\right|\ge0\forall y\\ \left|z+\dfrac{1}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\forall x,y,z\\ \)

Mà \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{2}{5}\right|=0\\\left|z+\dfrac{1}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{2}{5}=0\\z+\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{2}{5}\\z=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy ...

c,

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

Mà

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-1890}{1975}=\dfrac{-378}{395}\\z=2004\end{matrix}\right. \)

Vậy ...

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....

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

\(\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|\ge0\\\left|y+\dfrac{4}{3}\right|\ge0\\\left|z+\dfrac{7}{2}\right|\ge0\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\end{matrix}\right.\)

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

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

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\\left|y-\dfrac{2}{5}\right|\ge0\\\left|z+\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}\right|=0\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}\right|=0\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|\ge0\\ \left|y+\dfrac{1980}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{matrix}\right.\)

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\Rightarrow x=-\dfrac{19}{5}\\ \left|y+\dfrac{1980}{1975}\right|=0\Rightarrow y=-\dfrac{1980}{1975}\\\left|z-2004\right|=0\Rightarrow z=2004\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\ \left|y-\dfrac{1}{5}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\)

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

Dấu "=" xảy ra khi:

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

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

\(\Rightarrow A\ge0\)

Mà ĐK đề là \(A\le0\)

\(\Rightarrow A=0\)

\(\left[{}\begin{matrix}\left|x+\dfrac{3}{4}=0\right|\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}=0\right|\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}=0\right|\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

Các câu còn lại tương tự nhé

bài 1:

a, x + |2 - x| = 6

=> |2 - x| = 6 - x (1)

=>\(\orbr{\begin{cases}2-x=6-x\\2-x=x-6\end{cases}}\Rightarrow\orbr{\begin{cases}2=6\left(ktm\right)\\x=4\left(tm\right)\end{cases}}\)

b. |x - 7| = 7 

=> \(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}\Rightarrow\orbr{\begin{cases}x=14\left(ktm\right)\\x=0\left(tm\right)\end{cases}}}\)

c, Tương tự b

bài 2:

a, Vì \(\hept{\begin{cases}\left|x+2\right|\ge0\\\left|y+5\right|\ge0\end{cases}}\forall x,y\Rightarrow\left|x+2\right|+\left|y+5\right|\ge0\) (1)

Mà |x + 2| + |y + 5| = 0 (2)

Từ (1),(2) => \(\hept{\begin{cases}x+2=0\\y+5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-5\end{cases}}\)

b, tương tự a

1)

a) x + | 2 - x | = 6

\(\Rightarrow\)| 2 - x | = 6 - x

\(\Rightarrow\)\(\orbr{\begin{cases}2-x=6-x\\2-x=x-6\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}2=6\\x=4\end{cases}}\)

b) | x - 7 | = 7

x - 7 = +;- 7

\(\Rightarrow\)\(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=14\\x=0\end{cases}}\)

c) | x + 1 | = 5

x + 1 = +;- 5

\(\Rightarrow\)\(\orbr{\begin{cases}x+1=5\\x+1=-5\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)

2) Tự làm :v

khong dua nao tra loi a . tra loi di chi cho 5 h