Tìm x, y thuộc Z biết:

a) x.y=11

b) (x+1).(y+3) = 6

c) 1+2+3+....+x= 55

d) (x+22) chia hết cho (x+1)

Giúp mình với ạ, mình đang cần gấp

Đọc tiếp...

X + \(\frac{5}{2}\)x  \(\frac{2}{3}\)=  \(\frac{5}{2}\)

X  +     \(\frac{5}{3}\)         =  \(\frac{5}{2}\)

           X                 =  \(\frac{5}{2}\)-  \(\frac{5}{3}\)

            X                =    \(\frac{5}{6}\)

~ Hok T ~

x + 5/2 x 2/3 = 5/2

x + 10/6        = 5/2

x                   = 5/2 - 10/6

x                    = 5/6

Ai trên 10 SP k tui ikkkkkkkk

=>(x-2023)[(x-2023)^21-1]=0

=>x-2023=0 hoặc x-2023=1

=>x=2023 hoặc x=2024

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

=>\(\dfrac{3}{x-5}-\dfrac{y}{3}=\dfrac{1}{6}\)

=>\(\dfrac{9-y\left(x-5\right)}{3\left(x-5\right)}=\dfrac{1}{6}\)

=>9-y(x-5)=1/2(x-5)

=>(x-5)(1/2+y)=9

=>(x-5)(2y+1)=18

=>\(\left(x-5;2y+1\right)\in\left\{\left(18;1\right);\left(-18;-1\right);\left(2;9\right);\left(-2;-9\right);\left(6;3\right);\left(-6;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(23;0\right);\left(-13;-1\right);\left(7;4\right);\left(3;-5\right);\left(11;1\right);\left(-1;-2\right)\right\}\)

\(x:2=x:3\)

\(\Rightarrow x=0\)

\(x:2=x:3\)

\(\Leftrightarrow\left(x:2\right)-\left(x:3\right)=0\)

\(\Leftrightarrow\left(x.\frac{1}{2}\right)-\left(x.\frac{1}{3}\right)=0\)

\(\Leftrightarrow x.\left(\frac{1}{2}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x.\frac{1}{6}=0\)

\(\Leftrightarrow x=0\)

X = 0 nhá