=>(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: \(x\ne\dfrac{1}{2}\))

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

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

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

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

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

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

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

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

cc đầu b***

\(2x^4-x^3+2x^2+1=2x^4-2x^3+2x^2+x^3-x^2+x+x^2-x+1\\ \)

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

Vậy a = 2; b = 1; c = 1.

Làm rõ hơn đi bạn

4x² = 100

=> x² = 100 : 4 

=> x² = 25

=> x² = 5²

=> x = 5

Trả lời:

\(\text{a) 48-(2x+9)=21+x }\)

\(48-2x-9=21+x\)

\(-2x-x=21+9-48\)

\(-3x=-18\)

\(x=6\)

Vậy x = 6

b)\(\left(x+4\right)\left(11-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+4=0\\11-x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x=11\end{cases}}\)

Vậy x = -4; x = 11

c)\(4x^2=100\)

\(x^2=25\)

\(x^2=5^2\)

\(\Rightarrow x=5\)

Vậy x = 5

~ Học tốt

#Huyền Anh =P

a) 48 - ( 2x + 9 ) =  21 + x
    48 - 2x - 9 = 21 + x
    -2x - x = 21 - 48 + 9
    -3x = -18
     -x = -6
      x = 6
b) \(\left(x+4\right)\left(11-x\right)=0 \)
\(\Rightarrow\orbr{\begin{cases}x+4=0\Rightarrow x=-4\\11-x=0\Rightarrow x=11\end{cases}}\)
c) \(4x^2=100\)
    \(x^2=100:4\)
    \(x^2=25\)
    \(\Rightarrow x=\pm5\)
    

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.

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

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

a, => x^3 < 0 ; x-3 > 0 hoặc x^3 > 0 ; x-3 < 0

=> 0 < x < 3

b, => x^4.(2x-8) < 0

=> x^4.(x-4) < 0

Vì x^4 >= 0

=> x-4 < 0

=> x  < 4

c, Vì x-1 < x+12

=> x-1 < 0 ; x+12 >0

=> -12 < x < 1

d, => x-12 > 0 ; x-1 > 0 hoặc x-12 < 0 ; x-1 < 0

=> x  >12 hoặc x < 1

Tk mk nha

Thank you so much

a, A lớn nhất khi 7x la nguyên dương nho nhất

\(\Rightarrow7x=1\)

\(\Rightarrow x=\frac{1}{7}\)

\(b,B=\frac{10+4-x}{4-x}\)

\(B=\frac{10}{4-x}+1\)

b lon nhat khi 4-xla nguyen duong nho nhat

\(\Rightarrow4-x=1\)

\(\Rightarrow x=4-1=3\)

\(c,C=\frac{27-2x}{12-x}=\frac{3+24-2x}{12-x}=\frac{3}{12-x}+2\)

c lon nhat khi 12-x la nguyen duong nho nhat

\(\Rightarrow12-x=1\Rightarrow x=11\)

a)x=1

b)x=3

c)x=11