ta có |x+19|+|y-5|+1980 >1980

<=>|x+19|+|y-5|>0

dấu"="chỉ xảy ra <=>|x+19|=0vs|y-5|=0<=>x+19=0vsy-5=0

                                   <=>x=-19,y=5

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

Bài 1 :

a)x.(x+3)=0

=>  x=0 hoặc x+3=0

ta có: x+3=0

          x   = -3

Vậy x=0 hoặc x=-3

b) (x-2). (5-x) = 0

=> x-2=0 hoặc 5-x =0

TH1   

x-2=0

x   =2

TH2

5-x  =0

  x   =5

Vậy x=5 hoặc x=2

Bài 2

a) Để A có GTNN thì | x: 9| + |y-5| < 0

=> A=1890 +|x:9|+ | y-5| < 1890

Dấu = chỉ xảy ra khi | x: 9|+|y-5|=0

a) 4 cặp

b) -10

a: \(\left|x+\frac{19}{55}\right|\ge0\forall x\)

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

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

Do đó: \(\left|x+\frac{19}{55}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}x+\frac{19}{55}=0\\ y+\frac{1890}{1975}=0\\ z-2004=0\end{cases}\Rightarrow\begin{cases}x=-\frac{19}{55}\\ y=-\frac{1890}{1975}=-\frac{378}{395}\\ z=2004\end{cases}\)

b: Sửa đề: \(\left|x+\frac92\right|+\left|y+\frac43\right|+\left|z+\frac72\right|\le0\)

Ta có: \(\left|x+\frac92\right|\ge0\forall x\)

\(\left|y+\frac43\right|>=0\forall y\)

\(\left|z+\frac72\right|\ge0\forall z\)

Do đó: \(\left|x+\frac92\right|+\left|y+\frac43\right|+\left|z+\frac72\right|\ge0\forall x,y,z\)

mà \(\left|x+\frac92\right|+\left|y+\frac43\right|+\left|z+\frac72\right|\le0\)

nên \(\begin{cases}x+\frac92=0\\ y+\frac43=0\\ z+\frac72=0\end{cases}\Rightarrow\begin{cases}x=-\frac92\\ y=-\frac43\\ z=-\frac72\end{cases}\)

c: \(\left|x+\frac34\right|\ge0\forall x\)

\(\left|y-\frac15\right|\ge0\forall y\)

\(\left|x+y+z\right|\ge0\forall x,y,z\)

Do đó: \(\left|x+\frac34\right|+\left|y-\frac15\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}x+\frac34=0\\ y-\frac15=0\\ x+y+z=0\end{cases}\Rightarrow\begin{cases}x=-\frac34\\ y=\frac15\\ z=-x-y=\frac34-\frac15=\frac{11}{20}\end{cases}\)

d: \(\left|x+\frac34\right|\ge0\forall x\)

\(\left|y-\frac25\right|\ge0\forall y\)

\(\left|z+\frac12\right|\ge0\forall z\)

Do đó: \(\left|x+\frac34\right|+\left|y-\frac25\right|+\left|z+\frac12\right|\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\begin{cases}x+\frac34=0\\ y-\frac25=0\\ z+\frac12=0\end{cases}\Rightarrow\begin{cases}x=-\frac34\\ y=\frac25\\ z=-\frac12\end{cases}\)