a, x= 3

b, ko có x thỏa mãn

c, x= 3

d, x= 2

a: x=3

b: \(2x-1=2\)

hay \(x=\dfrac{3}{2}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

a: \(\Leftrightarrow x\in\left\{10;-10\right\}\)

b: \(\Leftrightarrow2x^2+4x-6x-12-3=0\)

\(\Leftrightarrow2x^2+2x-15=0\)

\(\Delta=2^2-4\cdot2\cdot\left(-15\right)=4+120=124\)

=>Ko có số nguyên x nào thỏa mãn bài toán

c: \(\Leftrightarrow2x-1\in\left\{1;-1;23;-23\right\}\)

hay \(x\in\left\{1;0;12;-11\right\}\)

a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)

\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)

b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)

\(\Leftrightarrow x=\frac{-3}{2}\)

c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)

\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)

d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)

\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)

lắm thế :)) ko hết đc :)) 

\(x+3⋮x+1\)

\(x+1+2⋮x+1\)

\(2⋮x+1\)

\(x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Tự lập bảng 

\(2x+5⋮x+2\)

\(2\left(x+2\right)+1⋮x+2\)

\(1⋮x+2\)

\(x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)

Tự lập bảng 

Bợn tự lm nốt đi :v 

b: =>12-x=-7

hay x=19