Đề ko rõ ràng \(\sqrt{x^2}+x+\dfrac{1}{4}\) hay \(\sqrt{x^2+x+\dfrac{1}{4}}\)??

m??

Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$

3.

$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$

$=(x-1)(x^2+3x+1)$

=>(2x-1)^2=24^2

=>2x-1=24 hoặc 2x-1=-24

=>x=-23/2 hoặc x=25/2

\(\dfrac{2x-1}{-12}=\dfrac{48}{1-2x}\) (ĐK: \(x\ne\dfrac{1}{2}\))

\(\Leftrightarrow\dfrac{2x-1}{12}=\dfrac{48}{2x-1}\)

\(\Leftrightarrow-12\cdot48=\left(2x-1\right)\left(2x-1\right)\)

\(\Leftrightarrow567=\left(2x-1\right)^2\)

\(\Leftrightarrow24^2=\left(2x-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=-24\\2x-1=24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-23\\2x=25\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(tm\right)\\x=\dfrac{25}{2}\left(tm\right)\end{matrix}\right.\)

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x^2-2x}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-2=x^2-2x\)

\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)

Cho mình sửa lại nhé:

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-2=x-2\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

\(\left(2x-1\right)^2+\left(x-3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};\dfrac{4}{3}\right\}\)

Chỗ cuối là (2x + 11) nha bạn
(2x + 1) + (2x + 3) + (2x + 5) +...+ (2x + 11) = 126
2x + 1 + 2x + 3 + 2x + 5 +...+ 2x + 11 = 126
(6 . 2x ) + (1 + 3 + 5 +...+ 11) = 126
Số số hạng của tổng (1 + 3 + 5 +...+ 11) là:
          (11 - 1) : 2 + 1 = 6 (số)
Tổng của (1 + 3 + 5 +...+ 11) là:
          (11 + 1) . 6 : 2 = 36
\(\Rightarrow\)12x + 36 = 126
          12x = 126 - 36
          12x = 90
\(\Rightarrow\) x = 90 : 12
           x = 7,5
Vậy x = 7,5