Tìm x,y nguyên:
1) /2x+1/+/y-4/< hoặc bằng 1
2) /3x/+/4+5/< hoặc bằng 2
3) 5./x+1/+/y-2/< hoặc bằng 7
4) 3./x+5/+2./y-1/< hoặc bằng 3
5)x-y=3 và /x/+/y/=3
6) x-2y=5 và /x/+/2y-1/=6
ai làm được tớ tick cho nhưng phải làm đầy đủ
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a, 2I3xI+Iy+3I=10 <=>6IxI+Iy+3I=10
vì 6IxI<=10 =>IxI<=10/6 <=>IxI<=1 => x=1;-1;0
x=1 hoặc x=-1=>Iy+3I=4 =>y=1 hoặc -7
x=0 => Iy+3I=10=>y=7 hoặc -13
b, Tương tự 12IxI<=21=>IxI<=21/12 =>IxI=1
x=1 hoặc -1 =>y=6 hoặc -12
x=0 => y= 18 hoặc -24
c, Tương tự I2x+1I<=3 <=> -3<= 2x+1<=3 <=>-4<= 2x<= 2 <=>-2<= x <=1
x=-2 hoặc 1=>Iy-4I=0 => y=4
x=-1 hoặc 0 =>Iy-4I=2 =>y=6 hoặc 2
d,2y^2+I2x+1I=5
tương tự 2y^2<=5 =>y^2<=5/2 <=>y^2<=2 =>y^2=1 hoặc 0
y^2=0 =>y=o thì I2x+1I=5 => x=2 hoặc -3
y^2=1 => y= 1 hoặc -1 thì I2x+1I=3 =>x =1 hoặc -2
Bài 1 :
\(3x+5=2\left(x-\frac{1}{4}\right)\)
\(\Leftrightarrow3x+5=2x-\frac{1}{2}\)
\(\Leftrightarrow5+\frac{1}{2}=2x-3x\)
\(\Leftrightarrow\frac{11}{2}=-x\)
\(\Leftrightarrow\frac{-11}{2}=x\)
Vậy \(x=\frac{-11}{2}\)
Bài 2:
a, \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{2018}{2019}\right|\ge0\\\left|z-3\right|\ge0\end{cases}}\)
Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)
\(\Rightarrow+,\left|x+\frac{19}{5}\right|=0\)
\(\Leftrightarrow x+\frac{19}{5}=0\)
\(\Leftrightarrow x=\frac{-19}{5}\)
\(\Rightarrow+,\left|y+\frac{2018}{2019}\right|=0\)
\(\Leftrightarrow y+\frac{2018}{2019}=0\)
\(\Leftrightarrow y=\frac{-2018}{2019}\)
\(\Rightarrow+,\left|z-3\right|=0\)
\(\Leftrightarrow z-3=0\)
\(\Leftrightarrow z=3\)
Vậy \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-2018}{2019}\\z=3\end{cases}}\)
b, Ta có : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)
Vì : \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y+4\right|\ge0\\\left|z-5\right|\ge0\end{cases}}\)
Mà : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)
\(\Rightarrow+,\left|x-\frac{1}{2}\right|\ge0\)
\(\Rightarrow x\inℚ\)
\(\Rightarrow+,\left|2y+4\right|\ge0\)
\(\Rightarrow y\inℚ\)
\(\Rightarrow+,\left|z-5\right|\ge0\)
\(\Rightarrow z\inℚ\)
Vậy chỉ cần \(\hept{\begin{cases}x\inℚ\\y\inℚ\\z\inℚ\end{cases}}\)thì thỏa mãn.
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!