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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.
\(\left(3-\frac{9}{10}-\left|x+2\right|\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1\)
\(\left(\frac{21}{10}-\left|x+2\right|\right):\frac{1}{2}+\frac{4}{5}=1\)
\(\left(\frac{21}{10}-\left|x+2\right|\right):\frac{1}{2}=\frac{1}{5}\)
\(\frac{21}{10}-\left|x+2\right|=\frac{1}{10}\)
=> |x+2| = 2
TH1: x + 2 = 2 => x = 0
TH2: x + 2 = -2 => x = -4
KL:...
\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
\(\left(-3-x\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-3-x=0\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)
x/3=y/5=x+y/3+5=16/8=2
x/3=2 suy ra x=6
y/5=2 suy ra y=10
x/2=y/3suy ra x/8=y/12
y/4=z/5 suy ra y/12=z/15
x/8=y/12=z/15=x+y-z/8+12-15=10/5=2
x/8=2 suy ra x=16
y/12=2 suy ra y=24
x/15=2 suy ra z=30
3 . 5x+2 + 4 . 5x-3 = 15 . 510
=> 5x-3 . ( 3.55+4) = 15.510
=> 5x-3 . 3129 = 3.511
=> 5x-1 . 1043 =511
=> 512-x = 1043
Số lẻ quá bạn ơi