tìm \(x\in Z\)
\(1+\frac{-1}{60}+\frac{19}{120}<\frac{x}{36}+\frac{-1}{60}<\frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
Tìm x thuộc Z, biết: \(1+\frac{-1}{60}+\frac{19}{120}
tìm x thuộc z biết 1 +\(\frac{1}{60}\)+ \(\frac{19}{120}< \frac{x}{36}+\frac{-1}{60}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
Tìm x\(\in\) Z biết 1+\(\frac{-1}{60}\)+\(\frac{19}{120}\)<\(\frac{x}{36}\)+\(\frac{-1}{60}<\frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
Tim x biet
\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
=> \(\frac{137}{120}< \frac{x}{36}< \frac{521}{360}\)
=> \(\frac{411}{360}< \frac{10x}{360}< \frac{521}{360}\)
=> 411 < 10x < 521
=> x \(\in\){ 42,43,44,...,52}
tìm x thuộc z biết 1 + 1/60 + 19/120 < x/36 + -1/60 < 58/90 + 59/72 + -1/60
Tìm x thuộc z ,biết:
1+ -1/60+19/120<x/36+ -1/60<58/90+59/72+ -1/60
Tìm x
a) \(1+\frac{-1}{60}\)\(+\frac{19}{120}\)<\(\frac{x}{36}\)\(+\frac{-1}{60}\)<\(\frac{58}{90}\)\(+\frac{59}{72}\)\(+\frac{-1}{60}\)
Tìm \(x,y,z\in Q\) biết:
a) \(|x+\frac{19}{5}|+|y+\frac{18}{19}|+|z-2004|=0\)
b) \(|x+\frac{3}{4}|+|y-\frac{1}{5}|+|x+y+z|=0\)
ta thấy /x+19/5/>=0
/y+18/19/>=0
/x-2004/>=0
Mà /x+19/5/+/y+18/19/+/z-2004/=0
=> x+19/5=0=>x=-19/5
y+18/19=0=>y=-18/19
z-2004=0=>z=2004
Câu còn lại tương tự nha bạn
Tích mik nha
b, \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
vì \(\left|x+\frac{3}{4}\right|\ge0\forall x;\left|y-\frac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall z\)
Dâu ''='' xảy ra <=> x = -3/4 ; y = 1/5 ; \(-\frac{3}{4}+\frac{1}{5}+z=0\Leftrightarrow z=\frac{11}{20}\)
\(1\)+\(\dfrac{-1}{60}+\dfrac{19}{120}< \dfrac{x}{36}< \dfrac{58}{90}+\dfrac{59}{72}+\dfrac{-1}{60}\)\(\left(x\in Z\right)\)
Mẫu số chung : \(LCM\left(60;120;36;90;72\right)=360\)
Quy đồng mẫu số :
\(\dfrac{360}{360}+\dfrac{-6}{360}+\dfrac{57}{360}< \dfrac{10\cdot x}{360}< \dfrac{232}{360}+\dfrac{295}{360}+\dfrac{-6}{360}\)
\(\Leftrightarrow\dfrac{411}{360}< \dfrac{10\cdot x}{360}< \dfrac{521}{360}\)
Vậy tập hợp các giá trị của x là \(x=\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)