a: =>ab-3b-a=5

=>a(b-1)-3b+3=8

=>(b-1)(a-3)=8

=>\(\left(a-3;b-1\right)\in\left\{\left(1;8\right);\left(8;1\right);\left(-1;-8\right);\left(-8;-1\right);\left(2;4\right);\left(4;2\right);\left(-2;-4\right);\left(-4;-2\right)\right\}\)

=>\(\left(a,b\right)\in\left\{\left(4;9\right);\left(11;2\right);\left(2;-7\right);\left(-5;0\right);\left(5;5\right);\left(7;3\right);\left(1;-3\right);\left(-1;-1\right)\right\}\)

b: =>ab-3b-3=5

=>b(a-3)=8

=>\(\left(a-3;b\right)\in\left\{\left(1;8\right);\left(8;1\right);\left(-1;-8\right);\left(-8;-1\right);\left(2;4\right);\left(4;2\right);\left(-2;-4\right);\left(-4;-2\right)\right\}\)

=>\(\left(a,b\right)\in\left\{\left(4;8\right);\left(11;1\right);\left(2;-8\right);\left(-5;-1\right);\left(5;4\right);\left(7;2\right);\left(1;-4\right);\left(-1;-2\right)\right\}\)

Bài 4: 

b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

a) \(\dfrac{4}{5}< \dfrac{5}{a}< \dfrac{10}{7}\) \(\left(a\inℤ\right)\)

\(\Leftrightarrow\dfrac{7}{10}< \dfrac{a}{5}< \dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{7.5}{10}< a< \dfrac{5}{4}.5\)

\(\Leftrightarrow\dfrac{7}{2}< a< \dfrac{25}{4}\)

\(\Leftrightarrow a\in\left\{4;5;6\right\}\)

b) \(\dfrac{2}{5}< \dfrac{a-1}{10}< \dfrac{8}{15}\)

\(\Leftrightarrow\dfrac{2.10}{5}< a-1< \dfrac{8.10}{15}\)

\(\Leftrightarrow4< a-1< \dfrac{16}{3}\)

\(\Leftrightarrow5< a< \dfrac{19}{3}\)

\(\Leftrightarrow a\in\left\{6\right\}\)

c) \(\dfrac{12}{7}< \dfrac{4}{a}< \dfrac{8}{3}\)

\(\Leftrightarrow\dfrac{3}{8}< \dfrac{a}{4}< \dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{3.4}{8}< a< \dfrac{7.4}{12}\)

\(\Leftrightarrow\dfrac{3}{2}< a< \dfrac{7}{3}\)

\(\Leftrightarrow a\in\left\{2\right\}\)

d) \(5< a^2-15< 16\)

\(\Leftrightarrow10< a^2< 31\)

\(\Leftrightarrow\sqrt[]{10}< a< \sqrt[]{31}\)

\(\Leftrightarrow a\in\left\{4;5\right\}\)

Ko chép đề

\(2)x-17+x=x-7\)

\(x+x-x=-7+17\)

\(x=-10\)

Vậy ....

âm 10 , nhỉ ?

b) Ta có: xy=-3

nên x,y là các ước của -3

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(1;-3\right);\left(-1;3\right);\left(-3;1\right);\left(3;-1\right)\right\}\)

Lời giải:

a. 

$(25-2x)^3:5-3^2=4^2$

$(25-2x)^3:5=4^2+3^2=25$

$(25-2x)^3=25.5=5^3$

$\Rightarrow 25-2x=5$

$\Rightarrow 2x=20$

$\Rightarrow x=10$

b.

$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$

$\Rightarrow 3^x=3^{14}$

$\Rightarrow x=14$