\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)

\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y=-1\\y=2\end{matrix}\right.\)

\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)

\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)

ai giúp tôi đi

\(\left(3x-5\right)⋮\left(x+2\right)\)

\(\Rightarrow3.\left(x+2\right)-11⋮\left(x+2\right)\)

Vì \(3.\left(x+2\right)⋮\left(x+2\right)\)

\(\Rightarrow11⋮\left(x+2\right)\)

\(\Rightarrow\left(x+2\right)\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Tự lập bảng :) T lười qá

\(\left(x+30\right)\left(2y+1\right)=14\)

\(\Rightarrow\left(x+30\right)\left(2y+1\right)=1.14=14.1=2.7=7.2=\left(-1\right)\left(-14\right)=\left(-14\right)\left(-1\right)=\left(-2\right)\left(-7\right)=\left(-7\right)\left(-2\right)\)Tự lập bảng và tìm giá trị của x, y :)

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.

234*(-26)+134*26

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