2(1-2x)-5=3(x+2)

=>\(2-4x-5=3x+6\)

=>\(-4x-3=3x+6\)

=>\(-7x=9\)

=>\(x=-\dfrac{9}{7}\)

Bài làm:

Ta có: \(\left(2x-5\right)\left(x^3+1\right)-\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)\left(x^2-x+1\right)-\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)\left(x^2-x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)\left(2x-5\right)=0\)

GPT này ra ta được: \(x\in\left\{-1;0;1;\frac{5}{2}\right\}\)

\(\left(2x-5\right)\left(x^3+1\right)-\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow2x^4+2x-5x^3-5-\left(2x^2+2x-5x-5\right)=0\)

\(\Leftrightarrow2x^4+2x-5x^3-5-2x^2-2x+5x+5=0\)

\(\Leftrightarrow2x^4+5x-5x^3-2x^2=0\)

\(\Leftrightarrow x\left(2x^3+5-5x^2-2x\right)=0\)

\(\Leftrightarrow x=0;\frac{5}{2};\pm1\)

\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)

\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)

\(\Leftrightarrow2x^2+5=0\)(vô lý)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)

\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)

Vậy pt vô nghiệm

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

Đề bài : Tìm số nguyên x.

|2x-1|+|2+x|+|x+3|=5(x-1)

2x-1+2+x+x+3=5x-5

2x+x+x-5x-1+2+3=-5

-x-1+5=-5

-x-1=(-5)-5

-x-1=-10

-x=(-10)+1

-x=-9

x=9

Vậy x=9.

Không chắc!

\(\left|2x-1\right|+\left|2+x\right|+\left|x+3\right|=5.\left(x-1\right)\left(1\right)\)

+)Ta có VT(1):\(\left|2x-1\right|\ge0;\left|2+x\right|\ge0;\left|x+3\right|\ge0\)

\(\Rightarrow\left|2x-1\right|+\left|2+x\right|+\left|x+3\right|\ge0\)

Mà VT(1)=VP(1)

\(\Rightarrow5.\left(x-1\right)\ge0\)

\(\Rightarrow x-1\ge0\)

\(\Rightarrow x\ge1\)

+)Ta lại có:\(x\ge1\Rightarrow2x-1\ge1\Rightarrow\left|2x-1\right|=2x-1\)(2)

                        \(x\ge1\Rightarrow2+x\ge3\Rightarrow\left|2+x\right|=2+x\)(3)

                       \(x\ge1\Rightarrow x+3\ge4\Rightarrow\left|x+3\right|=x+3\)(4)

+)Từ (2);(3) và (4) thì (1) trở thành:

2x-1+2+x+x+3=5.(x-1)

2x+x+x+2-1+3=5.(x-1)

4x+4              =5.(x-1)

4x+4              =5x-5

4+5                 =5x-4x

9                     =x

\(\Rightarrow\)x                  =9

Vậy x=9

Chúc bn học tốt

5^2\(x+1\)  + 3¹ = 128

5^\(2x+1\)  + 3 = 128

5^{\(^{2x+1}\)}     = 128 - 3

5\(^{2x+1}\)   = 125

5\(^{2x+1}\)   = 5³

2\(x\) + 1 = 3

2\(x\)       = 3 - 1

2\(x\)       =   2

 \(x\)        = 2: 2

 \(x\)        = 1