\(a,\dfrac{6}{5}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{6}{5}\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\\ \Rightarrow x=-21:\dfrac{3}{4}\\ \Rightarrow x=-28\\ c,\dfrac{2}{-7}=\dfrac{18}{x}\\ \Rightarrow x=18:\dfrac{2}{-7}\\ \Rightarrow x=-63\\ d,\dfrac{-5}{2}=\dfrac{10}{-x}\\ \Rightarrow x=-10:\dfrac{-5}{2}\\ \Rightarrow x=4\)

\(a,\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow6.x=5.18=90\\ \Rightarrow6.x=90\\ \Rightarrow x=15\\ b,\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow3.x=4.21=84\\ \Rightarrow x=28\)

Bài 4:

a) \(\dfrac{x}{2}=\dfrac{2}{4}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{1}{2}\)

\(\Rightarrow x=2\)

Vậy: \(x=2\)

b) \(-\dfrac{1}{5}=\dfrac{2}{x}\)

\(\Rightarrow x=\dfrac{-5.2}{1}=-10\)

Vậy: \(x=-10\)

c) \(\dfrac{x}{5}=\dfrac{5}{x}\)

\(\Rightarrow x^2=25\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{5;-5\right\}\)

\(\dfrac{x}{2}=\dfrac{2}{4}\\ =>x=\dfrac{2.2}{4}=1\)

\(\dfrac{-1}{5}=\dfrac{2}{x}\\ =>x=\dfrac{2.5}{-1}=-10\)

\(\dfrac{x}{5}=\dfrac{5}{x}\\ =>x^2=25\\ x=5;x=-5\)

4: \(\left(x^2-1\right)\left(x^2-6\right)<0\)

TH1: \(x^2-1>0\) và \(x^2-6<0\)

=>1<\(x^2\) <6

mà x là số nguyên

nên x∈{2;-2}

TH2: \(x^2-1<0\) và \(x^2-6>0\)

=>\(x^2<1\) và \(x^2>6\)

mà 6>1

nên x∈∅

Bài 2:Sửa đề: Tìm x,y nguyên

(2x+1)(5-y)=6

=>(2x+1)(y-5)=-6

mà 2x+1 lẻ

nên (2x+1;y-5)∈{(1;-6);(-1;6);(3;-2);(-3;2)}

=>(2x;y)∈{(0;-1);(-2;11);(2;3);(-4;7)}

=>(x;y)∈{(0;-1);(-1;11);(1;3);(-2;7)}

a: =>x+28=0

=>x=-28

b: =>27-x=0 hoặc x+9=0

=>x=27 hoặc x=-9

c: =>x=0 hoặc x-43=0

=>x=0 hoặc x=43

Bài 4:

a: |x-2|=0

=>x-2=0

=>x=2

b: |x+3|=1

=>\(\left[\begin{array}{l}x+3=1\\ x+3=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1-3=-2\\ x=-1-3=-4\end{array}\right.\)

c: \(-3\left|4-x\right|=-9\)

=>|4-x|=3

=>|x-4|=3

=>\(\left[\begin{array}{l}x-4=3\\ x-4=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=1\end{array}\right.\)

d: |2x+1|=-2

mà -2<0<=|2x+1|

nên x∈∅

Bài 5:

a: \(\left(x+3\right)^2=36\)

=>\(\left[\begin{array}{l}x+3=6\\ x+3=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6-3=3\\ x=-6-3=-9\end{array}\right.\)

b: \(\left(x+5\right)^2=100\)

=>\(\left[\begin{array}{l}x+5=10\\ x+5=-10\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10-5=5\\ x=-10-5=-15\end{array}\right.\)

c: \(\left(2x-4\right)^2=0\)

=>2x-4=0

=>2x=4

=>x=2

d: \(\left(x-1\right)^3=27\)

=>x-1=3

=>x=4

a: \(x\in\left\{1;7\right\}\)

b: \(x+1=1\)

hay x=0

\(a,\Rightarrow x\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ b,\Rightarrow2\left(x+1\right)-1⋮x+1\\ \Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Rightarrow x\in\left\{-2;0\right\}\)

a: =>y/15=-2/3

hay y=-10

b: 2/x=x/18

nên \(x^2=36\)

hay \(x\in\left\{6;-6\right\}\)

c: x/9=16/x

nên \(x^2=144\)

hay \(x\in\left\{12;-12\right\}\)

a)x=-10
b)x=-6

c)x=-12 nhé

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: =>x/-3=3

hay x=-9

b: =>x/9=-1/9

hay x=-1

c: =>x+1/5=-1/3

hay x=-8/15

d: =>-7/x=-7/9

hay x=9

a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)

b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)

c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)

d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)